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\(x=1+\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Rightarrow\left(x-1\right)^3=2+4+3\sqrt[3]{2.4}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)=6+6\left(x-1\right)=6x\)
\(\Rightarrow x^3-3x^2+3x-1=6x\Rightarrow x^3-3x^2-3x-1=0\)
Ta có:
\(M=\left(x^5-3x^4-3x^3-x^2\right)-x^4+4x^3-2x+2015\)
\(\Rightarrow M=x^2\left(x^3-3x^2-3x-1\right)-x^4+3x^3+3x^2+x+x^3-3x^2-3x-1+2016\)
\(\Rightarrow M=-x\left(x^3-3x^2-3x-1\right)+\left(x^3-3x^2-3x-1\right)+2016\)
\(\Rightarrow M=2016\)
1/ \(x-1=\sqrt[3]{2}\Rightarrow\left(x-1\right)^3=2\Rightarrow x^3-3x^2+3x-3=0\)
\(B=x^2\left(x^3-3x^2+3x-3\right)+x\left(x^3-3x^3+3x-3\right)+x^3-3x^2+3x-3+1945\)
\(B=1945\)
b/ Tương tự:
\(x-1=\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow x^3-3x^2+3x-1=6+3\sqrt[3]{8}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Rightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)
\(\Rightarrow x^3-3x^2-3x-1=0\)
\(P=x^2\left(x^3-3x^2-3x-1\right)-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x-1+2016\)
\(P=2016\)
f/
ĐKXĐ: ...
Đặt \(\sqrt{2-x}+\sqrt{x+2}=a>0\)
\(\Rightarrow a^2=4+2\sqrt{4-x^2}\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}\)
Phương trình trở thành:
\(a+\frac{a^2-4}{2}=2\)
\(\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}=0\)
\(\Rightarrow4-x^2=0\Rightarrow x=\pm2\)
e/ ĐKXĐ: ...
Đặt \(\sqrt{x+1}+\sqrt{4-x}=a>0\)
\(\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\)
Pt trở thành:
\(a+\frac{a^2-5}{2}=5\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=2\)
\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=4\)
\(\Leftrightarrow-x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a) Do VT >=0 nên VP >=0 nên \(x\ge4\)
\(PT\Leftrightarrow\left(x-2\right)-\sqrt{x-2}-2=0\)
Đặt \(\sqrt{x-2}=t\ge\sqrt{4-2}=\sqrt{2}\) thì \(t^2-t-2=0\)
\(\Leftrightarrow t=2\left(loại t = -1 vì nó không thỏa mãn đk\right)\Leftrightarrow x-2=4\Leftrightarrow x=6\)
Ta có \(x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
<=> \(\left(x-1\right)^3=6+3.\sqrt[3]{2.4}.\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
<=>\(x^3-3x^2+3x-1=6+6.\left(x-1\right)\)
<=>\(x^3-3x^2-3x-1=0\)
=> \(P=x^2\left(x^3-3x^2-3x-1\right)-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x-1+2016\)
=> \(P=2016\)