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NT
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16 tháng 1
a: \(x^3+8x=5x^2+4\)
=>\(x^3-5x^2+8x-4=0\)
=>\(x^3-x^2-4x^2+4x+4x-4=0\)
=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: \(x^3+3x^2=x+6\)
=>\(x^3+3x^2-x-6=0\)
=>\(x^3+2x^2+x^2+2x-3x-6=0\)
=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
3: ĐKXĐ: x>=0
\(2x+3\sqrt{x}=1\)
=>\(2x+3\sqrt{x}-1=0\)
=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)
=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)
=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)
=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)
4: \(x^4+4x^2+1=3x^3+3x\)
=>\(x^4-3x^3+4x^2-3x+1=0\)
=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)
=>(x-1)^2=0
=>x-1=0
=>x=1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1
a.
\(x^3+8x=5x^2+4\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
b.
\(x^3+3x^2-x-6=0\)
\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)
\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
3 tháng 2 2021
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
DT
3 tháng 2 2021
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}
25 tháng 11 2021
\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)
NB
1
VT
15 tháng 10 2019
dk \(x\ge0;2x+1\ge0< =>x\ge0\)
2(x+1)\(\sqrt{x}+\sqrt{3\left(x+1\right)^2\left(2x+1\right)}=\left(x+1\right)\left(5x^2-8x+8\right)< =>\)
\(2\sqrt{x}+\sqrt{3\left(2x+1\right)}=5x^2-8x+8\)(x+1>0 với x\(\ge0\)) <=>
2\(\sqrt{x}-2+\sqrt{6x+3}-3=5x^2-8x+3\) <=>\(\frac{2\left(x-1\right)}{\sqrt{x}+1}+\frac{6\left(x-1\right)}{\sqrt{6x+3}+3}=\left(x-1\right)\left(5x-3\right)< =>\)x-1=0 <=>x= 1 hoặc
\(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{6x+3}+3}=5x-3\)
x>1 thì \(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{6x+3}+3}< \frac{2}{1+1}+\frac{6}{3+3}=2\) hay 5x- 3<2 <=> x<1( vô lý)
x<1 thì \(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{6x+3}+}>2\) hay 5x-3>2 <=> x>1 (vô lý)
x=1 thỏa mãn
vậy pt có nghiệm duy nhất x=1
\(ĐK:x\ge\frac{-1}{3}\)
\(4x^3+5x^2+1=\sqrt{3x+1}-3x\Leftrightarrow4x^3+5x^2+3x+1-\sqrt{3x+1}=0\)\(\Leftrightarrow x\left(4x^2+5x+3\right)-\frac{3x}{\sqrt{3x+1}+1}=0\)\(\Leftrightarrow x\left(4x^2+5x+3-\frac{3}{\sqrt{3x+1}+1}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tmđk\right)\\4x^2+5x+3-\frac{3}{\sqrt{3x+1}+1}=0\end{cases}}\)
Xét phương trình \(4x^2+5x+3-\frac{3}{\sqrt{3x+1}+1}=0\)\(\Leftrightarrow\left(4x^2+5x+3\right)\sqrt{3x+1}+4x^2+5x=0\)\(\Leftrightarrow\left[\left(x+1\right)\left(4x+1\right)+2\right]\sqrt{3x+1}+4x^2+5x=0\)\(\Leftrightarrow\left(x+1\right)\left(4x+1\right)\sqrt{3x+1}+2\sqrt{3x+1}+4x^2+5x=0\)\(\Leftrightarrow\left(x+1\right)\left(4x+1\right)\sqrt{3x+1}+4x^2+x+4x+1+2\sqrt{3x+1}\)\(-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x+1\right)\sqrt{3x+1}+x\left(4x+1\right)+4x+1\)\(+\frac{12x+3}{2\sqrt{3x+1}+1}=0\)
\(\Leftrightarrow\left(4x+1\right)\left[\left(x+1\right)\sqrt{3x+1}+x+1+\frac{3}{2\sqrt{3x+1}+1}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\left(tmđk\right)\\\left(x+1\right)\sqrt{3x+1}+x+1+\frac{3}{2\sqrt{3x+1}+1}=0\end{cases}}\)
Với \(x\ge\frac{-1}{3}\)thì \(\left(x+1\right)\sqrt{3x+1}+x+1+\frac{3}{2\sqrt{3x+1}+1}>0\)
Vậy phương trình có tập nghiệm \(S=\left\{0;-\frac{1}{4}\right\}\)
ĐK: \(x\ge\frac{-1}{3}\)
\(4x^3+5x^2+1=\sqrt{3x+1}-3x\)
\(\Leftrightarrow4x^3+5x^2+1-\sqrt{3x+1}+3x=0\)
\(\Leftrightarrow4x^3+5x^2+1+\left(2x+1\right)-\sqrt{3x+1}=0\)
\(\Leftrightarrow4x^3+5x^2+x+\frac{\left(2x+1\right)^2-\left(3x+1\right)}{\left(2x+1\right)+\sqrt{3x+1}}=0\)
\(\Leftrightarrow\left(4x^2+x\right)\left(x+1\right)+\frac{4x^2+x}{\left(2x+1\right)+\sqrt{3x+1}}=0\)
\(\Leftrightarrow\left(4x^2+x\right)\left[\left(x+1\right)+\frac{1}{\left(2x+1\right)+\sqrt{3x+1}}\right]=0\)(*)
Với \(x\ge\frac{-1}{3}\)thì \(\left(x+1\right)+\frac{1}{\left(2x+1\right)+\sqrt{3x+1}}>0\)
(*) \(\Leftrightarrow4x^2+x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{4}\end{cases}\left(tmđk\right)}\)
Vậy phương trình có nghiệm \(x=0;x=\frac{-1}{4}\)