\(\Leftrightarrow x^2+2x+1+\left|x-1\right|=x^2+4\)

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

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

a)\(\dfrac{7x-1}{2}+2x=\dfrac{16-x}{3}\)

\(\dfrac{\left(7x-1\right).3}{2.3}+\dfrac{2x.6}{6}=\dfrac{\left(16-x\right)2}{3.2}\)

khử mẫu 

=> (7x-1).3+12x=(16-x).2

=>21x-3+12x=-2x+32

=>21x-3+12x+2x-32=0

=>35x-35=0

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

ĐKXĐ: x khác +-2

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

khử mẫu

(x+1).(x+2)+(x-1)(x-2)=2x²+4

=>x²+x+2+x+2+x²-2x-x+2=2x²+4

=>x²+x+2+x+2+x²-2x-x+2-2x²-4=0

=>(x²+x²-2x²)+(x+x-2x-x)+(2+2+2-4)=0

=>-x+2=0

=>-x=-2

=>x=2(loại)

vậy pt vô nghiệm

Ta có \(4\left(x-1\right)^2-\left(x+1\right)^2=x+13\Leftrightarrow4\left(x^3-3x^2+3x-1\right)-\left(x^2+2x+1\right)=x+13\)

\(\Leftrightarrow4x^3-12x^2+12x-4-x^2-2x-1-x-13=0\)

\(\Leftrightarrow4x^3-13x^2+9x-18=0\)\(\Leftrightarrow\left(4x^3-12x^2\right)-\left(x^2-3x\right)+\left(6x-18\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4x^2-x+6\right)=0\Leftrightarrow\orbr{\begin{cases}x-3=0\\4x^2-x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\4x^2-x+6=0\left(1\right)\end{cases}}}\)

Ta thấy (1) vô nghiệm vì \(\Delta=1-24=-23< 0\)

Vậy phương trình có nghiệm x=3

Ai trả lời giúp mình nhanh với

\(\frac{x-2}{x-4}-\frac{1}{x-2}=-2\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}-\frac{x-4}{\left(x-4\right)\left(x-2\right)}+2=0\)

\(\Leftrightarrow\frac{x^2-4x+4-x+4+2\left(x-4\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow x^2-5x+8+2x^2-12x+16=0\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)

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

\(\Leftrightarrow x=3\)hoặc \(x=\frac{8}{3}\)

\(\Rightarrow S=\left\{3;\frac{8}{3}\right\}\)

\(\frac{x-2}{x-4}-\frac{1}{x-2}=-2\)

\(< =>\frac{\left(x-2\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}-\frac{x-4}{\left(x-4\right)\left(x-2\right)}=-2\)

\(< =>\frac{x^2-2x-2x+4}{x^2-2x-4x+8}-\frac{x+4}{x^2-2x-4x+8}=\frac{-2\left(x^2-2x-4x+8\right)}{x^2-2x-4x+8}\)

\(< =>x^2-4x+4-x-4=-2x^2+4x+8x-16\)

\(< =>x^2-5x=-2x^2+12x-16\)

\(< =>x^2+2x^2-5x-12x+16=0\)

\(< =>3x^2-17x+16=0\)

Ta có : \(\Delta=\left(-17\right)^2-4.3.16=289-192=97\)

Nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{17+\sqrt{97}}{6}\)

\(x_2=\frac{17-\sqrt{97}}{6}\)

Vậy tập nghiệm của phương trình trên là \(\left\{\frac{17+\sqrt{97}}{6};\frac{17-\sqrt{97}}{6}\right\}\)

câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)

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

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

<=>(2x+1)(6-2x)=0

bước sau tự làm nốt nha !

câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a

Bài 2: 

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

\(\Leftrightarrow x-3=5\left(2x-3\right)=10x-15\)

=>-9x=-12

hay x=4/3

b: \(\Leftrightarrow x\left(x+2\right)-x+2=2\)

=>x²+2x-x+2=2

=>x²+x=0

=>x=0(loại) hoặc x=-1(nhận)

c: \(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

=>4=4(luôn đúng)

Vậy: S={x|x<>2; x<>-2}

\(\Leftrightarrow16-3\left(x+1\right)< 24+2\left(x-1\right)\)

=>16-3x-3<24+2x-2

=>-3x+13<2x+22

=>-5x<9

hay x>-9/5

\(\frac{x}{x+1}+\frac{x+1}{x+2}+\frac{x+2}{x}=\frac{25}{6}\)

<=> 6x²(x + 2) + 6x(x + 1)² + 6(x + 2)²(x + 1) = 25x(x + 1)(x + 2)

<=> 18x² + 54x² + 54x + 24 = 25x³ + 75x² + 50x

<=> 18x² + 54x² + 54x + 24 - 25x² - 75x² - 50x = 0

<=> -7x³ - 21x² + 4x + 24 = 0

<=> (-7x² - 28x - 24)(x - 1) = 0

vì 7x² + 28x + 24 khác 0 nên:

<=> x - 1 = 0

<=> x = 0

tìm ra đáp án chưa

Đc rồi chỉ mình với

Bài 3: 

a: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8=x^2-9\)

=>2x=6

hay x=3(loại)

b: \(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

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

=>2x=3

hay x=3/2