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a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)
Đặt \(t=x^2+6x+5\)
\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)
Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)
b) Đặt \(t=\left(2x+1\right)^2\)
\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)
Thay t:
\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)
a) \(x^2\left(x^2+4\right)-x^2-4=x^2\left(x^2+4\right)-\left(x^2+4\right)=\left(x^2+4\right)\left(x^2-1\right)=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
a. \(x^2\left(x^2+4\right)-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2-1\right)\left(x^2+4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+4\right)\)
b. \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)
Đặt \(t=x^2+7x+10\), ta được
(*) \(=t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)\)
hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)
Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo). Viết đề thế này khó đọc lắm.
\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)
Dấu ngoặc và cuối là sai nhé bạn. Phải là ngoặc vuông (x=0 hoặc x=-8) mới đúng, vì x không thể nhận 2 giá trị khác nhau cùng lúc.
=>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
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)
\(\Rightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}=\dfrac{2}{x^2-4}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=2\)
\(\Leftrightarrow x^2+3x+2=2\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) (thỏa mãn)
đkxđ: \(x ≠2; x ≠-2\)
\(\dfrac{x+1}{x-2}=\dfrac{2}{x^2-4}\)
\(⇔\dfrac{(x+1)(x+2)}{x^2-4}=\dfrac{2}{x^2-4}\)
\(⇔(x+1)(x+2)=2\)
\(⇔x^2+3x=0\)
\(⇔x(x+3)=0\)
\(⇔\left[\begin{array}{} x=0\\ x+3=0 \end{array} \right.\)
\(⇔\left[\begin{array}{} x=0\\ x=-3 \end{array} \right.\)
\(x\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x=0\) hay \(x+3=0\) hay \(x^2+1=0\) (pt vô nghiệm vì \(x^2+1\ge1\))
\(\Leftrightarrow x=0\) hay \(x=-3\)
-Vậy \(S= \left\{0;-3\right\}\)
\(x.\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-1-24=0\)
\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)
\(\Leftrightarrow x^2+x-1=\pm5\)
\(TH1:x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(TH2:x^2+x-1=-5\)
\(\Leftrightarrow x^2+x+4=0\)
Vì \(x^2+x+4=x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\)
Mà \(x^2+x+4=0\)
=> pt vô nghiệm
Vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
x( x - 1 )( x + 1 )( x + 2 ) = 24
<=> [ x( x + 1 ) ][ ( x - 1 )( x + 2 ) ] = 24
<=> ( x2 + x )( x2 + x - 2 ) - 24 = 0
Đặt t = x2 + x
pt <=> t( t - 2 ) - 24 = 0
<=> t2 - 2t - 24 = 0
<=> t2 - 6t + 4t - 24 = 0
<=> t( t - 6 ) + 4( t - 6 ) = 0
<=> ( t - 6 )( t + 4 ) = 0
<=> ( x2 + x - 6 )( x2 + x + 4 ) = 0
<=> ( x2 - 2x + 3x - 6 )( x2 + x + 4 ) = 0
<=> [ x( x - 2 ) + 3( x - 2 ) ]( x2 + x + 4 ) = 0
<=> ( x - 2 )( x + 3 )( x2 + x + 4 ) = 0
Vì x2 + x + 4 = ( x + 1/2 )2 + 15/4 ≥ 15/4 > 0 ∀ x
=> x - 2 = 0 hoặc x + 3 = 0
<=> x = 2 hoặc x = -3
Vậy ...