Lời giải:

Đặt \((\sqrt{1+x}=a; \sqrt{1-x}=b)\)

\(\Rightarrow a^2+b^2=2\) và \(a^2-b^2=2x\)

Khi đó:

\(M=\frac{\sqrt{1+ab}(a^3-b^3)}{2+ab}=\frac{\sqrt{1+ab}(a-b)(a^2+ab+b^2)}{a^2+b^2+ab}\)

\(=\sqrt{1+ab}(a-b)\)

\(=\sqrt{\frac{a^2+b^2}{2}+ab}(a-b)=\sqrt{\frac{a^2+b^2+2ab}{2}}(a-b)\)

\(=\sqrt{\frac{(a+b)^2}{2}}(a-b)=\frac{(a+b)(a-b)}{\sqrt{2}}=\frac{a^2-b^2}{\sqrt{2}}=\frac{2x}{\sqrt{2}}=\sqrt{2}x\)

\(M=\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{2+\sqrt{1-x^2}}\)

\(\Leftrightarrow M=\dfrac{\sqrt{2}.\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)

\(\Leftrightarrow M=\dfrac{\sqrt{2+2\sqrt{1-x^2}}\left[(\sqrt{\left(1+x\right)})^3-(\sqrt{\left(1-x\right)})^3\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)

\(\Leftrightarrow M=\dfrac{\sqrt{\left(1-x\right)+2\sqrt{\left(1-x\right)\left(1+x\right)}+(1+x)}.\left[(\sqrt{1+x})^3-\left(\sqrt{1-x}\right)^3\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)

\(\Leftrightarrow M=\dfrac{\sqrt{(\sqrt{1+x}+\sqrt{1-x})^2}.\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(\sqrt{1+x}\right)^2+\sqrt{1+x}\sqrt{1-x}+\left(\sqrt{1-x}^2\right)\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)

\(\Leftrightarrow M=\dfrac{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[1+x+\sqrt{1-x^2}+1-x\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)

\(\Leftrightarrow M=\dfrac{(1+x-1+x)\left[2+\sqrt{1-x^2}\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)

\(\Leftrightarrow M=\dfrac{2x}{\sqrt{2}}\)

\(\Leftrightarrow M=\sqrt{2}x\)

Câu hỏi của Đẹp Trai Không Bao Giờ Sai - Toán lớp 7 | Học trực tuyến tương tự

Với mọi x,y ta có :

\(+,\left|x-2y-1\right|\ge0\)

+, \(\left|y-4\right|+2\ge2\Leftrightarrow\dfrac{10}{\left|x-4\right|+2}\le5\)

Dấu "=" xảy ra khi :

\(\left\{{}\begin{matrix}\left|x-2y-1\right|=5\\\dfrac{10}{\left|x-4\right|+2}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=9\end{matrix}\right.\)

Vậy ..

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

Đề có sai không bạn?