mk cx có câu ntn

a)  \(\left(\frac{2}{5}x-2\right)-\left(\frac{3}{2}x+1\right)-\left(-4x-\frac{4}{5}\right)=\)\(\frac{18}{5}\)

\(\frac{2}{5}x-2-\frac{3}{2}x-1-4x+\frac{4}{5}=\frac{18}{5}\)

\(\frac{2}{5}x-\frac{3}{2}x-4x=\frac{18}{5}+2+1-\frac{4}{5}\)

\(\frac{8}{20}x-\frac{30}{20}x-\frac{80}{20}x=\frac{14}{5}+3\)

\(\frac{-51}{10}x=\frac{14}{5}+\frac{15}{5}\)

\(\frac{-51}{10}x=\frac{29}{5}\)

\(x=\frac{29}{5}.\frac{-10}{51}\)

\(x=\frac{-58}{51}\)

vậy \(x=\frac{-58}{51}\)

x=-58/51

1a) (0,4x - 2) - (1,5x + 1) - (-4x - 0,8) = 3,6

=> 0,4x - 2 - 1,5x - 1 + 4x + 0,8 = 3,6

=> 0,4x - 1,5x + 4x - 2 - 1 + 0,8 = 3,6

=> 2,9x - 2,2 = 3,6

=> 2,9x = 3,6 + 2,2

=> 2,9x = 5,8

=> x = 2

b)\(\left(\dfrac{3}{4}x+5\right)-\left(\dfrac{2}{3}x-4\right)-\left(\dfrac{1}{6}x+1\right)=\left(\dfrac{1}{3}x+4\right)-\left(\dfrac{1}{3}x-3\right)\)

\(\Rightarrow\dfrac{3}{4}x+5-\dfrac{2}{3}x+4-\dfrac{1}{6}x-1=\dfrac{1}{3}x+4-\dfrac{1}{3}x+3\)

\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{2}{3}x-\dfrac{1}{6}x\right)+\left(5+4-1\right)=7\)

\(\Rightarrow-\dfrac{1}{12}x+8=7\)

\(\Rightarrow-\dfrac{1}{12}x=-1\)

\(\Rightarrow x=12\)

a)\(1,2-x+0,8=-1,8-2x\)

\(2-x=-1,8-2x\)

\(2x-x=-1,8-2\)

\(x=-3,8\)

Vậy S={-3,8}

b)\(2,3x-1,4-4x=3,6-1,7x\)

\(2,3x-4x+1,7x=3,6+1,4\)

0=5(vô lí)

Vậy S={\(\varnothing\)}

c)\(6,6-0.9=2,6+0,1x-4\)

\(5,7=0,1x-1,4\)

\(-4,3=0,1x\)

\(x=-43\)

Câu c đáng lẽ là như thế này chứ.

c, 3(2.2-0.3x)=2.6+(0.1x-4)

<=> 6.6-0.9x=2.6+0.1x-4

<=> 6.6-0.9x=0.1x-1.4

<=>-0.9x -0.1x =-8

<=> -x=-8

<=> x=8

Mình trả lời câu d luôn nhé.

d, 3.6-0.5(2x+1)=x-0.25(2-4x)

<=> 3.6-x-0.5=x-0.5+x

<=> 3.1-x=2x-0.5

<=>-x-2x=-3.6

<=> -3x=-3.6

<=> x= 1.2

3,6 – 0,5(2x + 1) = x – 0,25(2 – 4x)

⇔ 3,6 – x – 0,5 = x – 0,5 + x ⇔ 3,6 – 0,5 + 0,5 = x + x + x

⇔ 3,6 = 3x ⇔ 1,2

Phương trình có nghiệm x = 1,2

=>3,6-x-0,5=x-0,5+2x

=>-4x=-0,5-3,1=-3,6

hay x=0,9

\(\Leftrightarrow3,6-x-0,5=x-0,5+x\Leftrightarrow-3x=-3,6\Leftrightarrow x=-1,2\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{2}{5}x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5:\dfrac{2}{5}=-\dfrac{25}{2}\end{matrix}\right.\)

Mấy câu như vậy bạn tải photomath về dùng nhé :)

e sẽ cố gắng !!! 

\(3x-15=2x\left(x-5\right)\)

\(3x-15=2x^2-10x\)

\(3x-15-2x^2+10x=0\)

\(13x-15-2x^2=0\)

\(x\left(13-2x\right)-15=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

\(f,x\left(2x-7\right)-4x+14=0\)

\(2x^2-7x-4x+14=0\)

\(2x^2-11x+14=0\)

\(x\left(2x-11\right)=-14\)

\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)

\(e,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)

\(3x+36-8x=x^2+4x+4\)

\(-5x+36-x^2-4x-4=0\)

\(-9x+32-x^2=0\)

\(x\left(-9-x\right)+32=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\23-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=23\end{cases}}}\)

chúc cj hay a hc tốt