\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

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

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)

giúp mik với

Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

Khó nhìn qá bn à

a: ta có: \(7-\left(2x-\frac13\right)^2=3\)

=>\(\left(2x-\frac13\right)^2=7-3=4\)

=>\(\left[\begin{array}{l}2x-\frac13=2\\ 2x-\frac13=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=2+\frac13=\frac73\\ 2x=-2+\frac13=-\frac53\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac73:2=\frac76\\ x=-\frac53:2=-\frac56\end{array}\right.\)

b: \(\left(2x+\frac13\right)^2-\frac38=\frac18\)

=>\(\left(2x+\frac13\right)^2=\frac38+\frac18=\frac48=\frac12\)

=>\(2x+\frac13=\sqrt{\frac12}=\frac{\sqrt2}{2}\)

=>\(2x=\frac{\sqrt2}{2}-\frac13=\frac{3\sqrt2-2}{6}\)

=>\(x=\frac{3\sqrt2-2}{12}\)

c: \(12:\left\lbrack29-\left(x-\frac23\right)^2\right\rbrack=3\)

=>\(29-\left(x-\frac23\right)^2=12:3=4\)

=>\(\left(x-\frac23\right)^2=29-4=25\)

=>\(\left[\begin{array}{l}x-\frac23=5\\ x-\frac23=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5+\frac23=\frac{17}{3}\\ x=-5+\frac23=-\frac{13}{3}\end{array}\right.\)

d: \(\left(3x-\frac12\right)^3+\frac83=\frac{29}{9}-\frac{14}{27}\)

=>\(\left(3x-\frac12\right)^3+\frac83=\frac{87}{27}-\frac{14}{27}=\frac{73}{27}\)

=>\(\left(3x-\frac12\right)^3=\frac{73}{27}-\frac83=\frac{73}{27}-\frac{72}{27}=\frac{1}{27}=\left(\frac13\right)^3\)

=>\(3x-\frac12=\frac13\)

=>\(3x=\frac12+\frac13=\frac56\)

=>\(x=\frac{5}{18}\)

e: \(2\left(2x-\frac13\right)^2+\frac43=\frac56+\frac{13}{18}\)

=>\(2\left(2x-\frac13\right)^2=\frac{15}{18}+\frac{13}{18}-\frac43=\frac{28}{18}-\frac{24}{18}=\frac{4}{18}=\frac29\)

=>\(\left(2x-\frac13\right)^2=\frac19\)

=>\(\left[\begin{array}{l}2x-\frac13=\frac13\\ 2x-\frac13=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac23\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=0\end{array}\right.\)