Đặt \(\left\{{}\begin{matrix}\sqrt[8]{1-x}=a\ge0\\\sqrt[8]{1+x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b+ab=3\\a^8+b^8=2\end{matrix}\right.\)

Ta có: \(a^8+7+b^8+7\ge8a+8b\)

\(a^8+b^8+6\ge8ab\)

\(\Rightarrow2\left(a^8+b^8\right)+20\ge8\left(ab+a+b\right)=24\)

\(\Rightarrow a^8+b^8\ge2\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=1\) hay \(x=0\)

1/ ĐKXĐ:...

\(\Leftrightarrow\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-2\sqrt{x+1}+1}=\frac{x+5}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(1-\sqrt{x+1}\right)^2}=\frac{x+5}{2}\)

\(\Leftrightarrow\sqrt{x+1}+1+\left|1-\sqrt{x+1}\right|=\frac{x+5}{2}\)

Nếu \(0\ge x\ge-1\Rightarrow\left|1-\sqrt{x+1}\right|=1-\sqrt{x+1}\)

\(\Rightarrow2=\frac{x+5}{2}\Leftrightarrow x=-1\left(tm\right)\)

Nếu \(x>0\Rightarrow\left|1-\sqrt{x+1}\right|=\sqrt{x+1}-1\)

\(\Rightarrow2\sqrt{x+1}=\frac{x+5}{2}\Leftrightarrow16x+16=x^2+10x+25\)

\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow x=3\left(tm\right)\)

Vậy...

Câu dưới tương tự

ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

\(\Leftrightarrow a^2-3a+2+ab-b=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)

\(\Leftrightarrow...\)

bằng 1 nữa nha

lớp 9 a

ĐK: \(x\ge0\)

\(PT\Leftrightarrow\frac{\sqrt{x+3}-\sqrt{x+2}}{1}+\frac{\sqrt{x+2}-\sqrt{x+1}}{1}+\frac{\sqrt{x+1}-\sqrt{x}}{1}=1\)

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)

\(\Leftrightarrow x+3+x-2\sqrt{x^2+3x}=1\)\(\Leftrightarrow2x+2=2\sqrt{x^2+3x}\)

\(\Leftrightarrow x^2+2x+1=x^2+3x\)

\(\Leftrightarrow x=1\)

Vậy.........................

\(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)

\(\Rightarrow\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{\left(x+1\right)\left(x+2\right)}\)

\(\Rightarrow\sqrt[3]{x+1}-1-\sqrt[3]{x+1}.\sqrt[3]{x+2}+\sqrt[3]{x+2}=0\)

\(\Rightarrow\left(\sqrt[3]{x+1}-1\right)-\sqrt[3]{x+2}\left(\sqrt[3]{x+1}-1\right)=0\)

\(\Rightarrow\left(\sqrt[3]{x+1}-1\right)\left(1-\sqrt[3]{x+2}\right)=0\)

Th1 : \(\sqrt[3]{x+1}-1=0\Rightarrow\sqrt[3]{x+1}=1\)

\(\Rightarrow x+1=1\Rightarrow x=0\)

Th2 : \(\sqrt[3]{x+2}-1=0\Rightarrow\sqrt[3]{x+2}=1\)

\(\Rightarrow x+2=1\Rightarrow x=-1\)

Vậy \(x\in\left\{0;-1\right\}\)

Đặt a = √(1-x)

b = √x

=> a² + b² = 1 và 1 + 2ab/3 = a + b

Giải hệ này tìm được a,b thế vô tìm được x

\(ĐK:-5\le x\le3\)

Đặt \(\sqrt{x+5}+\sqrt{3-x}=t\ge0\Leftrightarrow t^2-8=2\sqrt{15-2x-x^2}\), PTTT:

\(t-t^2+8-2=0\\ \Leftrightarrow t^2-t-6=0\\ \Leftrightarrow t=3\left(t\ge0\right)\\ \Leftrightarrow2\sqrt{15-2x-x^2}=3^2-8=1\\ \Leftrightarrow60-8x-4x^2=1\\ \Leftrightarrow4x^2+8x-59=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+3\sqrt{7}}{2}\left(tm\right)\\x=\dfrac{-2-3\sqrt{7}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm pt là ...

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