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a/
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-1}=a\\\sqrt[3]{27-14x}=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}2a+b=1\\14a^3+b^3=13\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-2a\\14a^3+b^3=13\end{matrix}\right.\)
\(\Rightarrow14a^3+\left(1-2a\right)^3=13\)
\(\Leftrightarrow a^3+2a^2-a-2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=-8\end{matrix}\right.\) \(\Leftrightarrow...\)
b/ ĐKXĐ: ...
\(VT=\sqrt{x-2}+\sqrt{4-x}\le\sqrt{2\left(x-2+4-x\right)}=2\)
\(VP=\left(x-3\right)^2+2\ge2\)
Đẳng thức xảy ra khi và chỉ khi \(x=3\)
Em thử nha,sai thì thôi ạ.
2/ ĐK: \(-2\le x\le2\)
PT \(\Leftrightarrow\sqrt{2x+4}-\sqrt{8-4x}=\frac{6x-4}{\sqrt{x^2+4}}\)
Nhân liên hợp zô: với chú ý rằng \(\sqrt{2x+4}+\sqrt{8-4x}>0\) với mọi x thỏa mãn đk
PT \(\Leftrightarrow\frac{6x-4}{\sqrt{2x+4}+\sqrt{8-4x}}-\frac{6x-4}{\sqrt{x^2+4}}=0\)
\(\Leftrightarrow\left(6x-4\right)\left(\frac{1}{\sqrt{2x+4}+\sqrt{8-4x}}-\frac{1}{\sqrt{x^2+4}}\right)=0\)
Tới đây thì em chịu chỗ xử lí cái ngoặc to rồi..
1.\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)
ĐK \(x\ge-1\)
Nhân liên hợp ta có
\(\left(x+3-x-1\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
<=>\(x^2+\sqrt{\left(x+1\right)\left(x+3\right)}=x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
<=> \(\left(x^2-x\sqrt{x+3}\right)+\left(\sqrt{\left(x+1\right)\left(x+3\right)}-x\sqrt{x+1}\right)=0\)
<=> \(\left(x-\sqrt{x+3}\right)\left(x-\sqrt{x+1}\right)=0\)
<=> \(\orbr{\begin{cases}x=\sqrt{x+3}\\x=\sqrt{x+1}\end{cases}}\)
=> \(x\in\left\{\frac{1+\sqrt{13}}{2};\frac{1+\sqrt{5}}{2}\right\}\)
Vậy \(x\in\left\{\frac{1+\sqrt{13}}{2};\frac{1+\sqrt{5}}{2}\right\}\)
\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)
\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)
\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)
\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)
\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)
\(\Leftrightarrow2\sqrt{x-8}+16=x\)
\(\Leftrightarrow x=24\)
\(a.\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{2}\sqrt{2+\sqrt{3}}.\)
\(=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3+1}\right)^2}\)
\(=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)^2=\left(2-\sqrt{3}\right)\left(4+2\sqrt{3}\right)\)
\(=2\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2\left(2^2-\sqrt{3}^2\right)=2\)
\(1.A=x-3\sqrt{x}+5=\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\) Điều kiện: \(x\ge0\)
\(\Rightarrow MinA=\frac{11}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\left(TM\right)\)
\(2.B=\left(x-2015\right)-\sqrt{x-2015}+2015=\left(\sqrt{x-2015}-\frac{1}{2}\right)^2+2015-\frac{1}{4}\) điều kiện: \(x\ge2015\)
\(B\ge2015-\frac{1}{4}=\frac{8059}{8060}\)
Dấu "=" xảy ra khi \(\sqrt{x-2015}-\frac{1}{2}=0\Leftrightarrow x-2015=\frac{1}{2^2}\Leftrightarrow x=\frac{8061}{8060}\left(TM\right)\)
\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)
\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
-3x+3=\(2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)
Chắc tới đây bạn làm đc rồi nhỉ
\(\left(x-2\right)^2\ge0;\forall x\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+5\ge5\\\left(x-2\right)^2+9\ge9\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x-2\right)^2+5}+\sqrt{\left(x-2\right)^2+9}\ge\sqrt{5}+\sqrt{9}=\sqrt{5}+3\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
\(A=\sqrt{x^2-8x+16}+\sqrt{x^2-24x+144}\)
\(A=\sqrt{\left(x-4\right)^2}+\sqrt{\left(12-x\right)^2}\)
\(A=\left|x-4\right|+\left|12-x\right|\ge\left|x-4+12-x\right|=8\)
\(A_{min}=8\) khi \(\left(x-4\right)\left(12-x\right)\ge0\Leftrightarrow4\le x\le12\)
ĐKXĐ: \(x\ge-2\)
- Với \(-2\le x< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}>1\Rightarrow\sqrt{x^2+1}-x>1\\\sqrt{x+3}\ge1\Rightarrow\sqrt{x+2}+\sqrt{x+3}\ge1\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x+2}+\sqrt{x+3}\right)>1\) pt vô nghiệm
- Với \(x\ge0\)
\(\Leftrightarrow\frac{1}{\sqrt{x^2+1}+x}\left(\sqrt{x+2}+\sqrt{x+3}\right)=1\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{x+3}=x+\sqrt{x^2+1}\)
\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+3}+x-\sqrt{x+2}=0\)
\(\Leftrightarrow\frac{x^2-x-2}{\sqrt{x^2+1}+\sqrt{x+3}}+\frac{x^2-x-2}{x+\sqrt{x+2}}=0\)
\(\Leftrightarrow\left(x^2-x-2\right)\left(\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+\frac{1}{x+\sqrt{x+2}}\right)=0\)
\(\Leftrightarrow x^2-x-2=0\Leftrightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
Dạ em cảm ơn Anh ạ