à, phần a ra x = 400. Nhầm

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2.\left(2x+3\right)-\left(4x+5\right)}{2.\left(5x+2\right)-\left(10x+2\right)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)

Suy ra:

\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow2.\left(2x+3\right)=1.\left(5x+2\right)\Rightarrow4x+6=5x+2\)

\(\Rightarrow x=4\)

a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow4x+30x-25x-8x=10-6\)

\(\Leftrightarrow1x=4\)

\(\Leftrightarrow x=4:1\)

\(\Leftrightarrow x=4\)

b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=966:138\)

\(\Leftrightarrow x=7\)

a. \(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(20x^2+34x+5=20x^2+33x+10\)

rút gọn \(x=5\)

mình gửi câu a tí nữa mình gửi câu b

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

1) 30x-30x^2-31

2)6x^4-2x^3-15x^2+23x-6

a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)

=-4/3x^2+8/3-10/3

=-4/3x^2-2/3

d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)

\(=3x^2+9x+22+\dfrac{68}{x-3}\)

a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)

\(\Rightarrow-80x=-480\Rightarrow x=6\)

b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)

\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)

\(\Rightarrow-4x=0\Rightarrow x=0\)

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow10x-12x-12x=-16+11+16-15\)

\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)

\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)

\(\Rightarrow5x-12x+24x-90x+36=182\)

\(\Rightarrow-73x=182-36\)

\(\Rightarrow-73x=146\Rightarrow x=-2\)

Chúc bạn học tốt!!!

a.

\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

b.

\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)

c.

\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d.

\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

e.

\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)

f.

\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)