Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

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

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6

a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

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

\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)

b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)

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

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

c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)

d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

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

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

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.

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

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

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

sửa lại chút: a) (2x+1)^2-2x-1=2              b) (x^2-3x)^2+5(x^2-3x)+6=0                  c) (x^2-x-1)(x^2-x)-2=0            d) (5-2x)^2+4x-10=8                   e) (x^2+2x+3)(x^2+2x+1)=3              f) x(x-1)(x^2-x+1)-6=0

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

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

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

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

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

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

Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)

Nếu ol thì tham khảo nah nguoiemtinhthong.

1.1

2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1

⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)

Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0

pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0

a=2ba=2b v a=13ba=13b

Các bạn tự giải quyết tiếp nhé.

1.2

TXĐ D=[1;+∞)D=[1;+∞)

đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0

pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0

⇔a=b⇔a=b v a=23ba=23b

...

1.3

D=[3;+∞)D=[3;+∞)

Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0

pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2

⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0

⇒a=5b⇒a=5b
...

1.4

ĐK

⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)

⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)

Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)

⇔2a2+2b2=3ab

1.5

Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)

⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x

⇔t2−t−4x2+2x=0t2−t−4x2+2x=0

Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2

⇒t=1−2xt=1−2x hoặc t=2xt=2x

1.1

2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1

2(.2+x+1)+3(x-1)

3a+b=11a2-19b2

tóm tắt

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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:

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