Câu hỏi toán lớp 7

I

iceghost

Bài 2

$\frac{x+1}{65}+ \frac{x+3}{63}= \frac{x+5}{61}+ \frac{x+7}{59}$

$\frac{x+1}{65}+ \frac{x+3}{63}+2= \frac{x+5}{61}+ \frac{x+7}{59}+2$

$\frac{x+1}{65}+ 1+ \frac{x+3}{63}+1= \frac{x+5}{61}+1+ \frac{x+7}{59}+1$

$\frac{x+66}{65}+ \frac{x+66}{63}= \frac{x+66}{61}+\frac{x+66}{59}$

$\frac{x+66}{65}+\frac{x+66}{63} - (\frac{x+66}{61}+\frac{x+66}{59})=0$

$\frac{x+66}{65}+\frac{x+66}{63} - \frac{x+66}{61}-\frac{x+66}{59}=0$

$(x+66).\frac{1}{65}+(x+66).\frac{1}{63}-(x+66).\frac{1}{61}-(x+66).\frac{1}{59}=0$ (Tỉ lệ thức)

$(x+66). (\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59})=0$

Ta thấy $\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59} \not= 0$ nên

$x+66 =0$
$x= -66$ :)
 
T

trucphuong02

[Toán 7]

1)
|2-3/2x|-4= x+2
\Rightarrow |2-3/2x|=x+6
\Rightarrow 2-3/2x= x+6 hoặc -x-6
TH1: 2-3/2x= x+6
\Rightarrow -3/2x-x=-2+6
\Rightarrow x.(-3/2-1)=4
\Rightarrow x.(-5/2)=4
\Rightarrow x=-8/5
TH2: 2-3/2x= -x-6
\Rightarrow -3/2x+x =-2-6
\Rightarrow x.(-3/2+1)=-8
\Rightarrow x.(-1/2)=8
\Rightarrow x= -16
Kết luận: x=-8/5
hoặc x= -16
 
T

truongtuan2001

Đã tìm ra

$| 2 - \frac{3}{2}x| - 4 = x+2$ (1)
TH1
$| 2 - \frac{3}{2}x|$ = $ 2 - \frac{3}{2}x $ khi $ 2 - \frac{3}{2}x $ \geq 0 \Rightarrow x \leq $\frac{4}{3}$

(1) \Rightarrow $ 2 - \frac{3}{2}x $ = x+6
\Leftrightarrow $\frac{-5}{2}$x = 4
\Leftrightarrow x = $\frac{-8}{5}$ (TM)

TH2

$| 2 - \frac{3}{2}x|$ = -2 + $\frac{3}{2}$x khi $ 2 - \frac{3}{2}x $ <0 \Leftrightarrow x< $\frac{4}{3}$
(1) \Rightarrow -2 + $\frac{3}{2}$x = x+6
\Leftrightarrow $\frac{1}{2}$x = 8
\Leftrightarrow x = 16(KTM)
Vậy tập nghiệm của ............................................. là.......
 
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