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a: Ta có: \(\sqrt{x^2-x+3}+7=10\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)
\(\Leftrightarrow x^2-4x+8=4\)
\(\Leftrightarrow x-2=0\)
hay x=2
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\(\Leftrightarrow\sqrt{x^3+3x^2+2x}=x^2-x-4\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=\left(x^2-x-4\right)^2\)
\(\Leftrightarrow x^3+3x^2+2x=x^4-2x^3-7x^2+8x+16\)
\(\Leftrightarrow-\left(x^2-2\right)\left(x^2-3x-8\right)\)
<=>-(x2-2)=0 hoặc x2-3x-8=0
Đối chiếu với đk ta thấy \(x=-\frac{\sqrt{41}-3}{2};\frac{\sqrt{41}+3}{2}\)thỏa mãn
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\(4x^2-5x-4\sqrt{x-1}-2=0\left(x\ge1\right)\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-\left(x-1+4\sqrt{x-1}+4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(\sqrt{x-1}+2\right)^2=0\)
\(\Leftrightarrow\left(2x-1-\sqrt{x-1}-2\right)\left(2x-1+\sqrt{x-1}+2\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x-1}-3\right)\left(2x+\sqrt{x-1}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=2x-3\\\sqrt{x-1}=-\left(2x+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x\in\varnothing\end{matrix}\right.\)
Vậy với x = 2 thì thỏa mãn pt
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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
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ĐKXĐ: ...
\(\Leftrightarrow3\left(2\sqrt{x+2}+\sqrt{3-x}\right)=3x+1+4\sqrt{-x^2+x+6}\)
Đặt \(2\sqrt{x+2}+\sqrt{3-x}=t>0\)
\(\Rightarrow t^2=4\left(x+2\right)+3-x+4\sqrt{\left(x+2\right)\left(3-x\right)}=3x+11+4\sqrt{-x^2+x+6}\)
Pt trở thành:
\(3t=t^2-10\)
\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2\sqrt{x+2}+\sqrt{3-x}=5\)
Ta có: \(VT=2\sqrt{x+2}+\sqrt{3-x}\le\sqrt{\left(2^2+1^2\right)\left(x+2+3-x\right)}=5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\frac{\sqrt{x+2}}{2}=\sqrt{3-x}\Leftrightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
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\(x^4-3x^3+4x^2-3x+1=0\)
Chia cả hai vế với \(x^2\)ta có
\(x^2-3x+4-\frac{3}{x}+\frac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-\left(3x+\frac{3}{x}\right)+4=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-3.\left(x+\frac{1}{x}\right)+4=0\)
Đặt \(t=x+\frac{1}{x}\left(t>0\right)\) \(\Rightarrow t^2-2=x^2+\frac{1}{x^2}\)
\(t^2-2-3t+4=0\)
\(\Leftrightarrow t^2-3t+2=0\)
\(\Leftrightarrow t^2-t-2t+2=0\)
\(\Leftrightarrow t.\left(t-1\right)-2.\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right).\left(t-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=1\left(TM\right)\\t=2\left(TM\right)\end{cases}}\)
TH1 \(t=1\)\(\Rightarrow x+\frac{1}{x}=1\)
\(\Leftrightarrow\frac{x^2+1}{x}=1\)\(\Leftrightarrow x^2+1=x\)
\(\Leftrightarrow x^2-x+1=0\)
\(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\) (Vô nghiệm)
TH2 \(t=2\) \(\Rightarrow x+\frac{1}{x}=2\)
\(\Leftrightarrow\frac{x^2+1}{x}=2\) \(\Leftrightarrow x^2+1=2x\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
(x+1)(x+2)(x+4)(x+8)=28x2
\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+9x+8\right)=28x^2\)(1)
Thấy x=0 không là nghiệm của (1). CHia 2 vế (1) cho x2 ta đc:
\(\left(1\right)\Leftrightarrow\left(x+\frac{8}{x}+6\right)\left(x+\frac{8}{9}+9\right)=28\)
Đặt \(t=x+\frac{8}{x}\)ta có:
\(\left(1\right)\Rightarrow\left(t+6\right)\left(t+9\right)=28\)
\(\Leftrightarrow t^2+15t+26=0\Leftrightarrow\orbr{\begin{cases}t=-2\\t=-13\end{cases}}\)
\(\Delta=13^2-4\left(1.8\right)=137\)\(\Rightarrow x_{1,2}=\frac{-13\pm\sqrt{137}}{2}\)(thỏa mãn)
Vậy...