gọi giao của (d₂) và (d₃) là A(x,y) suy ra x, y thỏa mãn hệ \(\hept{\begin{cases}2x-y=-1\\-3x-2y=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\Leftrightarrow A\left(-1.-1\right)\in\left(d\right)}\)

thay vào ta được \(-5m-6m+6=4m+6\Rightarrow m=0\)

vậy m=0 thỏa mãn đề bài

\(\frac{5x-1}{3}+\frac{7x-1,1}{3}-\frac{1,5-5x}{7}=\frac{9x-0,7}{4}\)

⇔ \(\frac{5x-1+7x-1,1}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

⇔ \(\frac{12x-2,1}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

⇔ \(\frac{28\left(12x-2,1\right)}{84}-\frac{12\left(1,5-5x\right)}{84}-\frac{21\left(9x-0,7\right)}{84}=0\)

⇔ \(\frac{336x-58,8}{84}-\frac{18-60x}{84}-\frac{189x-14,7}{84}=0\)

⇔ \(\frac{336x-58,8-18+60x-189x+14,7}{84}=0\)

⇔ \(\frac{207x-62,1}{84}=0\)

⇔ 207x - 62, 1 = 0

⇔ 207x = 62, 1

⇔ x = 0, 3

\(\frac{5x-1}{3}+\frac{7x-1.1}{3}-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\left(\frac{5x-1}{3}+\frac{7x-1.1}{3}\right)-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\left(\frac{5x-1+7x-1.1}{3}\right)-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\frac{12x-2.1}{3}-\frac{1.5-5x}{7}=\frac{9x-0,7}{4}\)

\(\Leftrightarrow\frac{28\left(12x-2.1\right)}{84}-\frac{12\left(1.5-5x\right)}{84}-\frac{21\left(8x-0,7\right)}{84}=0\)

\(\Leftrightarrow\frac{336x-58.8-18+60x-189x+14.7}{84}=0\)

\(\Leftrightarrow336x-58.8-18+60x-189x+14.7=0\)

\(\Leftrightarrow207x-62.1=0\)

\(\Leftrightarrow207x=62.1\)

\(\Leftrightarrow x=\frac{62.1}{207}=\frac{3}{10}=0.3\)

\(\frac{\left(5x-1\right)\left(7x-1,1\right)}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

\(\frac{35-5,5x-7x-11}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

\(\frac{24-12,5x}{3}-\frac{1,5-5x}{7}-\frac{9x-0,7}{4}=0\)

\(\frac{28.\left(24-12,5x\right)-12.\left(1,5-5x\right)-21\left(9x-0,7\right)}{84}=0\)

\(\frac{672-350x-18+60x-189x+14,7}{84}=0\)

\(\frac{668,7-479x}{84}=0\)

=> \(\left(668,7-479x\right).\frac{1}{84}=0\)

\(668,7-479x=0\)

\(479x=668,7\)

\(x=139,47\)

Bài mk ko biết có đúng hay ko nữa :((

Sai thì thôi nhé nhớ giúp mk nhé cảm ơn bạn nhìu

\(\frac{2\left(x-4\right)}{3}+\frac{4\left(x+3\right)-x+1}{8}=\frac{3\left(2x-3\right)}{5}-7\)

<=> \(\frac{2x-8}{3}+\frac{4x+12-x+1}{8}-\frac{6x-9}{5}+7=0\)

<=> \(\frac{40\left(2x-8\right)}{120}+\frac{15\left(3x+13\right)}{120}-\frac{24\left(6x-9\right)}{120}+\frac{840}{120}=0\)

<=> \(\frac{80x-320}{120}+\frac{45x+195}{120}-\frac{144x-216}{120}+\frac{840}{120}=0\)

<=> \(\frac{80x-320+45x+195-144x+216+840}{120}=0\)

<=> \(\frac{-19x+931}{120}=0\)

<=> -19x + 931 = 0

<=> -19x = -931

<=> x = 49

{=}-19x+931=0

{=}-19x=-931

{=}x=49

( x - 3 )³ - 2( x - 1 ) = x( x - 2 )² - 5x²

<=> x³ - 9x² + 27x - 27 - 2x + 2 = x( x² - 4x + 4 ) - 5x²

<=> x³ - 9x² + 25x - 25 = x³ - 4x² + 4x - 5x²

<=> x³ - 9x² + 25x - 25 - x³ + 4x² - 4x + 5x² = 0

<=> 21x - 25 = 0

<=> 21x = 25

<=> x = 25/21

\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)

\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x\left(x^2-4x+4\right)-5x^2\)

\(\Leftrightarrow x^3-9x^2+25x-25=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow x^3-9x^2+25x-25-x^3+4x^2-4x+5x^2=0\)

\(\Leftrightarrow21x-25=0\Leftrightarrow x=\frac{25}{21}\)

x = 38/11 nha

h đúng cho mk nha

Tham khảo câu b xem

https://olm.vn/hoi-dap/detail/248753766595.html

CHÚC BẠN HOK TỐT 

thui để tui đánh tay zậy

a, Xét tứ giác ABDC có: AM=MD (gt) ; BM=MC (gt)

=> ABDC là hình bình hành

b,Để ABDC là hình thoi => AB = AC => \(\Delta ABC\)cân

c, I đâu ra vậy bạn?