Toán 9 Giải phương trình

[quynh_anh06]

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[shorlochomevn@gmail.com]

[tex]\frac{}{}[/tex]

View attachment 142998
Giúp mình mấy câu này với ạ @shorlochomevn@gmail.com @Hoàng Vũ Nghị @Mộc Nhãn

7, đặt: [tex]\sqrt{x^3+x^2-2}=a[/tex]
pt trở thành: [tex]3a^2+x^2=2.(\frac{a^2+x^2}{x}).a\\\\ <=> x^3+3a^2x=2a^3+2ax^2\\\\ <=> x^3+3a^2x-2ax^2-2a^3=0\\\\ <=> x^3-ax^2-ax^2+a^2x+2a^2x-2a^3=0\\\\ <=> (x-a).(x^2-ax+2a^2)=0\\\\ <=> x=a\\\\ <=> x^3+x^2-2=x^2\\\\ <=>...[/tex]
8, chắc tương tự 7...
đặt [tex]\sqrt{11x-10}=a[/tex]
pt trở thành: [tex]4x^2+3a^2=7.\frac{a^2}{x}.a\\\\ <=> 4x^3+3a^2x=7a^3\\\\ <=> 4x^3-7a^3+3a^2x=0\\\\ <=> 4x^3-4a^2x+7a^2x-7a^3=0\\\\ <=> (x-a).(4x^2+7a^2)=0\\\\ <=> x=a\\\\ <=> x^2=11x-10\\\\ <=>...[/tex]
9, [tex] <=> 2x^3+ (16x^2-8x+1).\sqrt{8x-1}=17x.(8x-1)\\\\ <=> 2x^3+(16x^2-a^2).a=17x.a^2\\\\ <=> 2x^3 + 16x^2a-a^3-17xa^2=0\\\\ <=> 2x^3-2x^2a+18x^2a-18xa^2+xa^2-a^3=0\\\\ <=> (x-a).(2x^2+18xa+a^2)=0\\\\ <=> x=a[/tex]

 

[shorlochomevn@gmail.com]

11,[tex]<=> \frac{x^3+14}{x+2}-3=2\sqrt{\frac{x^3-3x+4}{x+1}}\\\\ <=> \frac{x^3-3x+8}{x+2}=2\sqrt{\frac{x^3-3x+4}{x+1}}\\\\ +, \sqrt{x^3-3x+4}=a; \sqrt{x+1}=b\\\\ pt <=> \frac{a^2+4}{b^2+1}=\frac{2a}{b}\\\\ <=> a^2b+4b=2ab^2+2a\\\\ <=> a^2b-2ab^2+4b-2a=0\\\\ <=> a.(ab-2)-2b.(ab-2)=0\\\\ <=> (a-2b).(ab-2)=0\\\\ +, a=2b <=> x^3-3x+4=x+1\\\\ <=> x^3-4x+3=0\\\\ <=> x^3-x^2+x^2-x-3x+3=0\\\\ <=> (x-1).(x^2+x-3)=0\\\\ <=>....\\\\ +, ab=2\\\\ <=> a^2b^2=4\\\\ <=> (x^3-3x+4).(x+1)=4\\\\ <=> x^4-3x^2+4x+x^3-3x+4=4\\\\ <=> x^4+x^3-3x^2+x=0\\\\ <=> x.(x^3+x^2-3x+1)=0\\\\ <=> x.(x^3-x^2+2x^2-2x-x+1)=0\\\\ <=>x.(x-1).(x^2+2x-1)=0\\\\ <=>...[/tex]
12, [tex]<=> \frac{a}{b}=\frac{2a^2.b+1}{2+b^3}\\\\ <=> 2a+ab^3=2a^2b^2+b\\\\ <=> 2a-b +ab^3-2a^2b^2=0\\\\ <=> (2a-b)-ab^2.(2a-b)=0\\\\ <=> (2a-b).(1-ab^2)=0\\\\ +, 2a-b=0 <=>....\\\\ +, 1-ab^2=0; 3b^2-2a^2=6x-3-6x+4=1\\\\ => b^2=\frac{1}{3}.(1+2a^2)\\\\ => a.\frac{1}{3}.(1+2a^2)=1\\\\ <=> 2a^3+a=3\\\\ <=> 2a^3+a-3=0\\\\ <=> 2a^3-2a^2+2a^2-2a+3a-3=0\\\\ <=> (a-1).(2a^2+2a+3)=0\\\\ <=> a=1 <=>...[/tex]
13, [tex]<=> 2ab+12=3a+8b\\\\ <=> 3a-12+8b-2ab=0\\\\ <=> 3.(a-4)-2b.(a-4)=0\\\\ <=> (a-4).(3-2b)=0\\\\ <=>...[/tex]
14,đặt: [tex]\sqrt{x}=a; \sqrt{x+1}=b;\sqrt{x+6}=c[/tex]
pt <=> [tex]<=> a+\frac{bc}{a}-b=c\\\\ <=> a^2+bc-ab-ac=0\\\\ <=> a.(a-b)-c.(a-b)=0\\\\ <=> (a-b).(a-c)=0\\\\ <=>...[/tex]
15,đặt: [tex]\sqrt{x}=a; \sqrt{x+2}=b; \sqrt{x+3}=c[/tex]
pt <=> [tex]\frac{1}{a}-ab=\frac{1}{c}-bc\\\\ <=> \frac{1}{a}-\frac{1}{c}-ab+bc=0\\\\ <=> \frac{c-a}{ac}+b.(c-a)=0\\\\ <=> (c-a).(\frac{1}{ac}+b)=0\\\\ <=> a=c[/tex]

 

[quynh_anh06]

[tex](x^2-6x+11)\sqrt{x^2-x+1}=2(x^2-4x+7)\sqrt{x-2}[/tex]

11,[tex]<=> \frac{x^3+14}{x+2}-3=2\sqrt{\frac{x^3-3x+4}{x+1}}\\\\ <=> \frac{x^3-3x+8}{x+2}=2\sqrt{\frac{x^3-3x+4}{x+1}}\\\\ +, \sqrt{x^3-3x+4}=a; \sqrt{x+1}=b\\\\ pt <=> \frac{a^2+4}{b^2+1}=\frac{2a}{b}\\\\ <=> a^2b+4b=2ab^2+2a\\\\ <=> a^2b-2ab^2+4b-2a=0\\\\ <=> a.(ab-2)-2b.(ab-2)=0\\\\ <=> (a-2b).(ab-2)=0\\\\ +, a=2b <=> x^3-3x+4=x+1\\\\ <=> x^3-4x+3=0\\\\ <=> x^3-x^2+x^2-x-3x+3=0\\\\ <=> (x-1).(x^2+x-3)=0\\\\ <=>....\\\\ +, ab=2\\\\ <=> a^2b^2=4\\\\ <=> (x^3-3x+4).(x+1)=4\\\\ <=> x^4-3x^2+4x+x^3-3x+4=4\\\\ <=> x^4+x^3-3x^2+x=0\\\\ <=> x.(x^3+x^2-3x+1)=0\\\\ <=> x.(x^3-x^2+2x^2-2x-x+1)=0\\\\ <=>x.(x-1).(x^2+2x-1)=0\\\\ <=>...[/tex]
12, [tex]<=> \frac{a}{b}=\frac{2a^2.b+1}{2+b^3}\\\\ <=> 2a+ab^3=2a^2b^2+b\\\\ <=> 2a-b +ab^3-2a^2b^2=0\\\\ <=> (2a-b)-ab^2.(2a-b)=0\\\\ <=> (2a-b).(1-ab^2)=0\\\\ +, 2a-b=0 <=>....\\\\ +, 1-ab^2=0; 3b^2-2a^2=6x-3-6x+4=1\\\\ => b^2=\frac{1}{3}.(1+2a^2)\\\\ => a.\frac{1}{3}.(1+2a^2)=1\\\\ <=> 2a^3+a=3\\\\ <=> 2a^3+a-3=0\\\\ <=> 2a^3-2a^2+2a^2-2a+3a-3=0\\\\ <=> (a-1).(2a^2+2a+3)=0\\\\ <=> a=1 <=>...[/tex]
13, [tex]<=> 2ab+12=3a+8b\\\\ <=> 3a-12+8b-2ab=0\\\\ <=> 3.(a-4)-2b.(a-4)=0\\\\ <=> (a-4).(3-2b)=0\\\\ <=>...[/tex]
14,đặt: [tex]\sqrt{x}=a; \sqrt{x+1}=b;\sqrt{x+6}=c[/tex]
pt <=> [tex]<=> a+\frac{bc}{a}-b=c\\\\ <=> a^2+bc-ab-ac=0\\\\ <=> a.(a-b)-c.(a-b)=0\\\\ <=> (a-b).(a-c)=0\\\\ <=>...[/tex]
15,đặt: [tex]\sqrt{x}=a; \sqrt{x+2}=b; \sqrt{x+3}=c[/tex]
pt <=> [tex]\frac{1}{a}-ab=\frac{1}{c}-bc\\\\ <=> \frac{1}{a}-\frac{1}{c}-ab+bc=0\\\\ <=> \frac{c-a}{ac}+b.(c-a)=0\\\\ <=> (c-a).(\frac{1}{ac}+b)=0\\\\ <=> a=c[/tex]

 

[shorlochomevn@gmail.com]

[tex](x^2-6x+11)\sqrt{x^2-x+1}=2(x^2-4x+7)\sqrt{x-2}[/tex]

[tex](x^2-6x+11)\sqrt{x^2-x+1}=2(x^2-4x+7)\sqrt{x-2}\\\\ đặt \sqrt{x^2-x+1}=a; \sqrt{x-2}=b\\\\ =>+, a^2-5b^2=x^2-x+1-5x+10=x^2-6x+11\\\\ +, a^2-3b^2=x^2-x+1-3x+6=x^2-4x+7\\\\ => pt <=> (a^2-5b^2).a=2(a^2-3b^2).b\\\\ <=> a^3-5ab^2=2a^2b-6b^3\\\\ <=> a^3-5ab^2-2a^2b+6b^3=0\\\\ <=> a^3-a^2b-a^2b+ab^2-6ab^2+6b^3=0\\\\ <=> (a-b).(a^2-ab-6b^2)=0\\\\ <=> (a-b).(a^2-3ab+2ab-6b^2)=0\\\\ <=> (a-b).(a-3b).(a+2b)=0\\\\ <=>....[/tex]

 

[Hanhh Mingg]

11,[tex]<=> \frac{x^3+14}{x+2}-3=2\sqrt{\frac{x^3-3x+4}{x+1}}\\\\ <=> \frac{x^3-3x+8}{x+2}=2\sqrt{\frac{x^3-3x+4}{x+1}}\\\\ +, \sqrt{x^3-3x+4}=a; \sqrt{x+1}=b\\\\ pt <=> \frac{a^2+4}{b^2+1}=\frac{2a}{b}\\\\ <=> a^2b+4b=2ab^2+2a\\\\ <=> a^2b-2ab^2+4b-2a=0\\\\ <=> a.(ab-2)-2b.(ab-2)=0\\\\ <=> (a-2b).(ab-2)=0\\\\ +, a=2b <=> x^3-3x+4=x+1\\\\ <=> x^3-4x+3=0\\\\ <=> x^3-x^2+x^2-x-3x+3=0\\\\ <=> (x-1).(x^2+x-3)=0\\\\ <=>....\\\\ +, ab=2\\\\ <=> a^2b^2=4\\\\ <=> (x^3-3x+4).(x+1)=4\\\\ <=> x^4-3x^2+4x+x^3-3x+4=4\\\\ <=> x^4+x^3-3x^2+x=0\\\\ <=> x.(x^3+x^2-3x+1)=0\\\\ <=> x.(x^3-x^2+2x^2-2x-x+1)=0\\\\ <=>x.(x-1).(x^2+2x-1)=0\\\\ <=>...[/tex]
12, [tex]<=> \frac{a}{b}=\frac{2a^2.b+1}{2+b^3}\\\\ <=> 2a+ab^3=2a^2b^2+b\\\\ <=> 2a-b +ab^3-2a^2b^2=0\\\\ <=> (2a-b)-ab^2.(2a-b)=0\\\\ <=> (2a-b).(1-ab^2)=0\\\\ +, 2a-b=0 <=>....\\\\ +, 1-ab^2=0; 3b^2-2a^2=6x-3-6x+4=1\\\\ => b^2=\frac{1}{3}.(1+2a^2)\\\\ => a.\frac{1}{3}.(1+2a^2)=1\\\\ <=> 2a^3+a=3\\\\ <=> 2a^3+a-3=0\\\\ <=> 2a^3-2a^2+2a^2-2a+3a-3=0\\\\ <=> (a-1).(2a^2+2a+3)=0\\\\ <=> a=1 <=>...[/tex]
13, [tex]<=> 2ab+12=3a+8b\\\\ <=> 3a-12+8b-2ab=0\\\\ <=> 3.(a-4)-2b.(a-4)=0\\\\ <=> (a-4).(3-2b)=0\\\\ <=>...[/tex]
14,đặt: [tex]\sqrt{x}=a; \sqrt{x+1}=b;\sqrt{x+6}=c[/tex]
pt <=> [tex]<=> a+\frac{bc}{a}-b=c\\\\ <=> a^2+bc-ab-ac=0\\\\ <=> a.(a-b)-c.(a-b)=0\\\\ <=> (a-b).(a-c)=0\\\\ <=>...[/tex]
15,đặt: [tex]\sqrt{x}=a; \sqrt{x+2}=b; \sqrt{x+3}=c[/tex]
pt <=> [tex]\frac{1}{a}-ab=\frac{1}{c}-bc\\\\ <=> \frac{1}{a}-\frac{1}{c}-ab+bc=0\\\\ <=> \frac{c-a}{ac}+b.(c-a)=0\\\\ <=> (c-a).(\frac{1}{ac}+b)=0\\\\ <=> a=c[/tex]

câu 10 thấy không hợp lí lắm vì nếu xét theo ĐKXĐ ý, thì không thể tách riêng 2 căn để đặt được đâu

 

[shorlochomevn@gmail.com]

câu 10 thấy không hợp lí lắm vì nếu xét theo ĐKXĐ ý, thì không thể tách riêng 2 căn để đặt được đâu

View attachment 142998
Giúp mình mấy câu này với ạ @shorlochomevn@gmail.com @Hoàng Vũ Nghị @Mộc Nhãn

10, [tex]<=> \sqrt{\frac{x^3+x-10}{x^3+x^2-3}}=\frac{6.(x^3+x-10)+5.(x^3+x^2-3)}{4.(x^3+x-10)+7.(x^3+x^2-3)}\\\\ <=> \sqrt{\frac{a}{b}}=\frac{6a+5b}{4a+7b}\\\\ => \frac{a}{b}=\frac{36a^2+25b^2+60ab}{16a^2+49b^2+56ab}\\\\ <=> 16a^3+49ab^2+56a^2b=36a^2b+25b^3+60ab^2\\\\ <=> 16a^3-11ab^2+20a^2b-25b^3=0\\\\ <=> 16a^3-16a^2b+36a^2b-36ab^2+25ab^2-25b^3=0\\\\ <=> (a-b).(16a^2+36ab+25b^2)=0\\\\ <=> (a-b).(4a+5b)^2=0\\\\ +, a=b <=>...\\\\ +, 4a=-5b \\\\ <=> 4.(x^3+x-10)=-5.(x^3+x^2-3)\\\\ <=> 4x^3+4x-40=-5x^3-5x^2+15\\\\ <=> 9x^3+5x^2+4x-55=0\\\\ <=> 5.(x^3+x^2-3)+4.(x^3+x-10)=0 (2)[/tex]
mà [tex]x^3+x^2-3[/tex] và [tex]x^3+x-10[/tex] cùng dấu và [tex]x^3+x^2-3\neq 0[/tex]
=> (2) vô nghiệm

 

[mbappe2k5]

câu 10 thấy không hợp lí lắm vì nếu xét theo ĐKXĐ ý, thì không thể tách riêng 2 căn để đặt được đâu

Nhưng mình đâu cần tìm hẳn ĐKXĐ, mình chỉ cần viết [TEX]{\frac{x^3+x-10}{x^3+x^2-3}}>0[/TEX] và [TEX]x^3+x^2-3\neq 0[/TEX] xong cuối cùng thử lại với giá trị [TEX]x[/TEX] đã tìm được là được thôi mà.