Kết quả tìm kiếm

$AB.AC=AH.BC=2S_{ABC}\Rightarrow S_{ABC}=\dfrac{1}{2}AB.AC\Rightarrow \triangle ABC$ vuông tại $A$

$\dfrac{1}{AB^2}=\dfrac{1}{AE^2}+\dfrac{1}{4AF^2} \\\Leftrightarrow \dfrac{4(AE^2-AB^2)}{AB^2.AE^2}=\dfrac{1}{AF^2} \\\Leftrightarrow \dfrac{4BE^2}{AB^2.AE^2}=\dfrac{1}{AF^2} \\\Leftrightarrow 4BE^2.AF^2=AB^2.AE^2 \\\Leftrightarrow 2BE.AF=2BC.AE \\\Leftrightarrow BE.AF=BC.AE$ $\Leftrightarrow...

$A=\left [ (\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}).\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y} \right ]\color{red}{:} \dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}$ bạn nhỉ? a) ĐK: $x>0;y>0$ $A=\left...

$(a^2+b^2)(x^2+y^2)=(ax+by)^2 \\\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2 \\\Leftrightarrow a^2y^2-2abxy+b^2x^2=0 \\\Leftrightarrow (ay-bx)^2=0 \\\Leftrightarrow ay=bx \\\Leftrightarrow \dfrac ax=\dfrac by$

Đề phải như này bạn nhỉ?

hình như cái này có vấn đề bạn ạ Nếu đặt $a=x-1$ thì $(a+2)^4+(a-2)^4=82$ bạn nhỉ? p/s: mk thấy 2 cách này cũng chẳng khác nhau Đặt $x-1=a\Rightarrow a=x+1$

$a) A=|x+2|+|x-6|=|x+2|+|6-x|\geq |x+2+6-x|=8$ Dấu '=' xảy ra khi $(x+2)(6-x)\geq 0\Leftrightarrow -2\leq x\leq 6$ Vậy... $b) B=|x+5|+|x+2|+|x-7|+|x-8| \\=|x+5|+|8-x|+|x+2|+|7-x| \\\geq |x+5+8-x|+|x+2+7-x| \\=13+9=22$ Dấu '=' xảy ra khi $\left\{\begin{matrix}(x+5)(8-x)\geq 0\\ (x+2)(7-x)\geq...

xem lại đề $E,K,F$ ko thẳng hàng bạn ạ

chắc viết nhầm, phải là $E$

thế kiểu gì bạn?

Rewrite the following sentences , using "UNLESS" 1.You will spoil it if you careful. You will spoil it unless you are careful 2.If you do not eat anything,you will be ill You will be ill unless you eat anything 3.You pass your examination if you do not work hard. You pass your examination...

$x=\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}=\dfrac{(\sqrt{7}+\sqrt{3})^2}{7-3}=\dfrac{10+2\sqrt{21}}{4}=\dfrac{5+\sqrt{21}}{2} \\y=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\dfrac{(\sqrt{7}-\sqrt{3})^2}{7-3}=\dfrac{10-2\sqrt{21}}{4}=\dfrac{5-\sqrt{21}}{2} \\\Rightarrow...

pt $\Leftrightarrow |x+1|.\left | 2x-\dfrac 34 \right |=2x-\dfrac14$ Nếu $x<-1$ thì $-(x+1)(\dfrac 34-2x)=2x-\dfrac14\Leftrightarrow x=\dfrac{3\pm \sqrt{73}}{16}$ (L) Nếu $-1\leq x<\dfrac 38$ thì $(x+1)(\dfrac 34-2x)=2x-\dfrac14\Leftrightarrow x=\dfrac{-13-3\sqrt{33}}{16}$ (L) or...

$\dfrac{\sqrt[3]{8-3\sqrt{5}}+\sqrt[3]{64-12\sqrt{20}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}} \\=\dfrac{\sqrt[3]{(8-3\sqrt{5})(8+3\sqrt{5})}+\sqrt[3]{8(8-3\sqrt{5})(8+3\sqrt 5)}}{\sqrt[3]{57}}...

gõ nhầm ^^. 6) ĐK: $x\geq \dfrac12$ pt $\Leftrightarrow \sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2$ $\Leftrightarrow \sqrt{(\sqrt{2x-1}+1)^2}+\sqrt{(\sqrt{2x-1}-1)^2}=2$ $\Leftrightarrow \sqrt{2x-1}+1+|\sqrt{2x-1}-1|=2$ Nếu $x>1$ thì $\sqrt{2x-1}=1\Leftrightarrow x=1$ Nếu $\dfrac12...

1) ĐK: $x\geq 4$ pt $\Leftrightarrow \sqrt{x+3}=1+\sqrt{x-4}$ $\Leftrightarrow x+3=1+x-4+2\sqrt{x-4}$ $\Leftrightarrow \sqrt{x-4}=0$ $\Leftrightarrow x=4$ (TM) 2) ĐK: $x\leq 3$ pt $\Leftrightarrow 15-x+3-x+2\sqrt{(15-x)(3-x)}=36$ $\Leftrightarrow \sqrt{(15-x)(3-x)}=x+9 \ (x\geq -9)$...

Died...

chắc là $A=(x+3y-5)^2-6xy+26$ $A=(x+3y-5)^2-6xy+26 \\=x^2+9y^2+25+6xy-10x-30y-6xy+26 \\=(x^2-10x+25)+(9y^2-30y+25)+1 \\=(x-5)^2+(3y-5)^2+1\geq 1$ Dấu '=' xảy ra khi $x=5;y=\dfrac 53$

$\triangla ABC$ vuông tại $A$ có $AI\perp BC$. Theo HTL ta có: $AB^2=BI.BC;AC^2=CI.BC\Rightarrow \dfrac{AB^2}{AC^2}=\dfrac{BI.BC}{CI.BC}=\dfrac{BI}{CI}$

$\triangle ABI$ vuông tại $I$ có $IM\perp AB\Rightarrow AI^2=AM.AB$ $\triangle ACI$ vuông tại $I$ có $IN\perp AC\Rightarrow AI^2=AN.AC$ $\Rightarrow AI^4=AB.AC.AM.AN$ Mà $AB.AC=BC.AI\Rightarrow AI^4=BC.AI.AM.AN\Rightarrow AI^3=BC.AM.AN$