=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

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

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

\(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\left(ĐKXĐ:x\ne1;x\ne-1\right)\) 

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

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

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

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

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

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy phương trình có nghiệm là x=0

mong có thể giúp bạn

f(x)g(x)=0<=>f(x)=0 hoặc g(x)=0

<=>(x²+x+1)²+x²+x-11=(x-1)(x+2)(x²+x+5) (=>vế phải có 3 TH)

TH1=>x=1

TH2=>x=-2

x²+x+5=0

1²-4(1.5)=-19

=>pt ko có nghiệm thực

=>x=1 hoặc -2

đặt x^2+x+1=a

a^2+a-12=0  <=>(a-3)(a+4)=0 

<=>a-3=0 hoặc a+4=0

Nếu a-3 =0 suy ra x^2+x-2=0 <=>(x+2)(x-1)=0 suy ra x=-2 hoặc x=1

Nếu a+4=0 thì x^2+x+5=0 vô nghiệm do x^2+x+1/4=(x+1/2)>=0

Lời giải:

$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$

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

$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
 

đk : x >= 0 ; x khác 4 

\(B=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right).\dfrac{\sqrt{x}-2}{2}=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(x-4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

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

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

\(=\left(x+2\right)\left(3x+5\right)-3x^2-12x-12=-x-2\)

Bạn nên dùng công thức trực quan cho bài toán như thế này nhé.

A = (2x - 1)(x + 2) - 3x² + (x - 1)²

= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1

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

= x - 1

B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)

= x³ - 8 - x³ - x² - 9 + x²

= (x³ - x³) + (-x² + x²) + (-8 - 9)

= -17

A = (2x - 1)(x + 2) - 3x² + (x - 1)²

= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1

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

= x - 1

B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)

= x³ - 8 - x³ - x² - 9 + x²

= (x³ - x³) + (-x² + x²) + (-8 - 9)

= -17

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