\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)

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

\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)

\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)

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

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

\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)

\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)

\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)

\(\Leftrightarrow x=1\left(koTM\right).\)

1) 4x(x-5)-(x-1)(4x-3)=5

<=>4x²-20x-4x²+3x+4x-3=5

<=>-13x=8

<=>x=-8/13

Thôi mỏi tay quá tìm x luôn nha

2) x=1.875

3) x=17/7

cho mình hỏi bạn làm kiểu gì vậy

b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

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

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)

\(\Leftrightarrow2x-8+12x=4x-2\)

\(\Leftrightarrow10x=6\)

hay \(x=\dfrac{3}{5}\)

2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)

\(\Leftrightarrow15x-6-30=10-20x\)

\(\Leftrightarrow35x=46\)

hay \(x=\dfrac{46}{35}\)

3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)

\(\Leftrightarrow3x-6-4=6x-6\)

\(\Leftrightarrow-3x=4\)

hay \(x=-\dfrac{4}{3}\)

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

\(\Leftrightarrow\dfrac{\left(x-4\right).2}{3.2}+\dfrac{2x.6}{6}=\dfrac{4x-2}{6}\)

\(\Rightarrow2x-8+12x=4x-2\\ \Leftrightarrow10x=6\\ \Leftrightarrow x=\dfrac{3}{5}\)

4: Ta có: \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)

\(\Leftrightarrow40x-20+45x-30=48x-36\)

\(\Leftrightarrow37x=14\)

hay \(x=\dfrac{14}{37}\)

5: Ta có: \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)

\(\Leftrightarrow2x-6-3x-6=x+4-9\)

\(\Leftrightarrow-x-x=-5-12=-17\)

hay \(x=\dfrac{17}{2}\)

1. Thu gọn biểu thức - Hoc24 làm rồi mà bạn?

1.

a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)

b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)

2.

a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)

b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

1,a,=x2−6x+8+3x2−15x=4x2−21x+8b,=9x2+12x+4−x2+9=8x2+12x+132,a,⇔x2+8x+16−x2+4=5⇔8x=−15⇔x=−158b,⇔9x2−6x+1−8x2−2x+12x+3−x2=5⇔4x=1⇔x=14

Sửa lại môn học để các bạn làm nhé em!

bạn sửa lại môn hôn học đi ạ

14) Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{2x+4}{x^2-2x-3}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x+8}{2\left(x-3\right)\left(x+1\right)}\)

Suy ra: \(x^2+x+x^2-3x-4x-8=0\)

\(\Leftrightarrow2x^2-6x-8=0\)

\(\Leftrightarrow x^2-3x-4=0\)

a=1; b=-3; c=-4

Vì a-b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=-1\left(loại\right);x_2=\dfrac{-c}{a}=4\left(nhận\right)\)