a) Ta có: \(3x\left(x+1\right)-2x\left(x+20\right)=-1-x\)

\(\Leftrightarrow3x^2+3x-2x^2-40x+1+x=0\)

\(\Leftrightarrow x^2-36x+1=0\)

\(\Leftrightarrow x^2-36x+324-323=0\)

\(\Leftrightarrow\left(x-18\right)^2=323\)

\(\Leftrightarrow\left[{}\begin{matrix}x-18=\sqrt{323}\\x-18=-\sqrt{323}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18+\sqrt{323}\\x=18-\sqrt{323}\end{matrix}\right.\)

Vậy: \(x\in\left\{18+\sqrt{323};18-\sqrt{323}\right\}\)

b) Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)-16=0\)

\(\Leftrightarrow6x^2+19x-7-6x^2-x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

hay x=1

Vậy: x=1

c) Ta có: \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)-8=0\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3-8=0\)

\(\Leftrightarrow-4x-5=0\)

\(\Leftrightarrow-4x=5\)

hay \(x=\frac{-5}{4}\)

Vậy: \(x=\frac{-5}{4}\)

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

a) x² - 5x - 6 = 0

=> x² - 2x - 3x - 6 = 0

=> (x² - 2x) + (-3x - 6) = 0

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

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

=> x - 2 = 0 => x = 2

     x - 3 = 0 => x = 3

còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546

Bạn giải giúp mình mấy câu kia được k=)))

\(\left(2x-1\right)^3-2x\left(2x+1\right)^2=5x-20x^2\)

\(\Rightarrow8x^3-12x^2+6x-1-2x\left(4x^2+4x+1\right)-5x+20x^2=0\)

\(\Rightarrow8x^3-12x^2+6x-1-8x^3-8x^2-2x-5x+20x^2=0\)

\(\Rightarrow-x-1=0\)

\(\Rightarrow x=-1\)

\(\left|5x+13\right|=2x-7\)

khi \(x>\frac{7}{2}\), biểu thức có dạng:

\(\orbr{\begin{cases}5x+13=2x-7\\5x+13=7-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=-20\\7x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{20}{3}\\x=-\frac{6}{7}\end{cases}}}\)

1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)

2 ) \(2x\left(b-a\right)+a-b=2x\left(b-a\right)-\left(b-a\right)=\left(2x-1\right)\left(b-a\right)\)

3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)

4 ) \(x^2-xy-2x+2y=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)

5 ) \(5x^2y+5xy^2+a^2x+a^2y\)

\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)

\(=\left(5xy+a^2\right)\left(x+y\right)\)

6 ) \(2x^2-6xy+5x-15y\)

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

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

7 ) \(ax^2-3axy+bx-3by\)

\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)

\(=x\left(ax+b\right)-3y\left(ax+b\right)\)

\(=\left(x-3y\right)\left(ax+b\right)\)

8 ) \(x^2+4x-5x-20=0\)

\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

9 ) \(x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

10 ) \(x^2-6x-4x+24=0\)

\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

:D

\(1) 2x+1=15-5x \)

\(⇔2x+5x=15-1\)

\(⇔7x=14\)

\(⇔x=2\)

vậy pt có 1 nghiệm là x=2

\(2) 3x-2=2x+5\)

\(⇔3x-2x=5+2\)

\(⇔x=7\)

vậy pt có 1 nghiệm là x=7

\(3) 7(x-2)=5(3x+1)\)

\(⇔7x-14=15x+5\)

\(⇔7x-15x=5+14\)

\(⇔-8x=19\)

\(⇔x=-\dfrac{19}{8}\)

vậy pt có 1 nghiệm là x=-\(\dfrac{19}{8}\)

\(4) 2x+5=20-3x\)

\(⇔2x+3x=20-5\)

\(⇔5x=15\)

\(⇔x=3\)

vậy pt có 1 nghiệm là x=3

\(5) -4x+8=0\)

\(⇔-4x=-8\)

\(⇔x=2\)

vậy pt có 1 nghiệm là x=2

\(6) x-3=10-5x\)

\(⇔x+5x=10+3\)

\(⇔6x=13\)

\(⇔x=\dfrac{13}{6}\)

vậy pt có 1 nghiệm là \(x=\dfrac{13}{6}\)

\(7) 3x-1=x+3\)

\(⇔3x-x=3+1\)

\(⇔2x=4\)

\(⇔x=2\)

vậy pt có 1 nghiệm là x=2

\(8) 2(x+1)=5x-7\)

\(⇔2x+2=5x-7\)

\(⇔2x-5x=-7-2\)

\(⇔-3x=-9\)

\(⇔x=3\)

vậy pt có 1 nghiệm là x=3.

1/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7-9\right)\left(2+\sqrt{x+3}\right)}{\left(4-x-3\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(2+\sqrt{x+3}\right)}{\left(x-1\right)\left(-\sqrt{2x+7}-3\right)}=\dfrac{2.4}{-6}=-\dfrac{4}{3}\)

2/ \(=\lim\limits_{x\rightarrow1^-}\dfrac{2.1-3}{1-1}=-\infty\)

3/ \(=\lim\limits_{x\rightarrow2^+}\dfrac{3-x}{x-2}=+\infty\)

4/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-\dfrac{8x^3}{x^2}+\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}-\dfrac{1}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-8x}{5}=\pm\infty\)

5/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}}+\dfrac{2x}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\dfrac{7}{x}}=\dfrac{1}{2}\)