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f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
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\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\\ \)(1)
\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)
\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0\Rightarrow!2x+1!=2x+1\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)
\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)
\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)
\(\left\{\begin{matrix}2x+1=0\\-x^2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-\frac{1}{2}\\x=0\end{matrix}\right.\)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2\left(x+\frac{1}{2}\right)\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)}=\left(x+\frac{1}{2}\right)\left(x^2+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}+1\right)}-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)}-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left(-1-x^2+1\right)=0\)
\(\Leftrightarrow-x^2\left(x+\frac{1}{2}\right)=0\)\(\Leftrightarrow\left[\begin{matrix}-x^2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=0\\x=-\frac{1}{2}\end{matrix}\right.\)
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a) \(x^2-5+\sqrt{x+5}=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\sqrt{x+5}=0\)(tự làm tiếp)
b) Đề hơi sai sai
c) Mik chưa nghĩ ra
d) \(\left(\sqrt{1-2x}-1\right)+\left(\sqrt{1+2x}-1\right)+x^2=0\)
\(\frac{-2x}{\sqrt{1-2x}+1}+\frac{2x}{\sqrt{1+2x}+1}+x^2=0\)(tự lm tiếp)
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Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ
Mn giúp em vs ạ! Thanks trước!
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Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
a) Ta có:
\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(\frac{\Leftrightarrow4}{x}-x+\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}=0\left(1\right)\)
Dật \(u=\sqrt{x-\frac{1}{x}};v=\sqrt{2x-\frac{5}{x}}\left(u,v\ge0\right)\Rightarrow u^2-v^2=\frac{4}{x}-x\)
Do đó (1) trở thành: \(u^2-v^2+u-v=0\Rightarrow u=v\)
Đến đây bạn tự giải nhé