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 ..

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

\(pt\Leftrightarrow\left|x+2\right|+\left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|=10x\)

Ta có: \(\left|x+2\right|+ \left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|\ge0\Leftrightarrow10x\ge0\Leftrightarrow x\ge0\)

Khi \(x\ge0\) thì: \(x+2+x+\dfrac{3}{5}+x+\dfrac{1}{2}=10x\)

\(\Rightarrow7x+2+\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{31}{10}\Leftrightarrow x=\dfrac{31}{70}\)

Ta có: \(\left|3x-5\right|+\left|3x+1\right|=\left|5-3x\right|+\left|3x+1\right|\ge\left|5-3x+3x+1\right|=6\)

Dấu "=" xảy ra \(\Leftrightarrow\left(5-3x\right)\left(3x+1\right)\ge0\Leftrightarrow-\dfrac{1}{3}\le x\le\dfrac{5}{3}\)

Vậy \(-\dfrac{1}{3}\le x\le\dfrac{5}{3}\)

\(\left(1\right)2xy\left(x-2y\right)+x-14y=0\)

\(\Leftrightarrow2xy\left(x-2y\right)+\left(x-2y\right)-12y=0\)

​\(\Leftrightarrow\left(2xy+1\right)\left(x-2y\right)=12y\)​

\(\left(2\right)xy\left(4xy+y+4\right)=y^2\left(2y+5\right)-1\)

\(\Leftrightarrow4x^2y^2+x^2y+4xy=2y^3+5y^2-1\)

\(\Leftrightarrow4x^2y^2+x^2y+4xy-2y^3-5y^2+1=0\)

\(\Leftrightarrow4x^2y^2+8xy+1-4xy+x^2+4y^2+x^2y-x^2-2y^3+2y^2-11y^2=0\)

\(\Leftrightarrow\left(2xy+1\right)^2+\left(x-2y\right)^2+x^2\left(y-1\right)-2y^2\left(y-1\right)=11y^2\)

\(\Leftrightarrow\left(2xy+1\right)^2+\left(x-2y\right)^2+\left(x^2-2y^2\right)\left(y-1\right)=11y^2\)

_ Phân tích được tới đây :)_

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có : \(2x+2y+z=4\)

\(\Rightarrow z=4-2x-2y\)

Khi đó \(A=2xy+yz+zx\)

\(=2xy+\left(y+x\right)z\)

\(=2xy+\left(y+x\right)\left(4-2x-2y\right)\)

\(=2xy+4y-2xy-2y^2+4x-2x^2-2xy\)

\(=4y+4x-2y^2-2x^2-2xy\)

\(\Rightarrow2A=-4x^2-4xy+8x-4y^2+8y\)

\(=-4x^2-4x\left(y-2\right)-4y^2+8y\)

\(=-4x^2-2.2x\left(y-2\right)-\left(y-2\right)^2+\left(y-2\right)^2-4y^2+8y\)

\(=-\left(2x+y-2\right)^2-3y^2+4y+4\)

\(=-\left(2x+y-2\right)^2-3\left(y^2-\dfrac{4}{3}y-\dfrac{4}{3}\right)\)

\(=-\left(2x+y-2\right)^2-3\left(y^2-2.y.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{4}{9}-\dfrac{4}{3}\right)\)

\(=-\left(2x+y-2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\le\dfrac{16}{3}\)

\(\Rightarrow A\le\dfrac{8}{3}\)

\(Max_A=\dfrac{8}{3}\Leftrightarrow\left\{{}\begin{matrix}y-\dfrac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{3}\\x=\dfrac{2}{3}\\z=\dfrac{4}{3}\end{matrix}\right.\)

Xét 2 th

x<3

x>=3

\(\left|2x-6\right|+5x=10\)

\(\Leftrightarrow\left|2x-6\right|=10-5x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=10-5x\\2x-6=5x-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5x=10+6\\-6+10=5x-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=16\\3x=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{16}{7}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy...

Mk cũng ở cái hoc24 này gần 2 năm rồi nhưng chưa ai nói mk xuống cấp đâu bạn à :)

\(\Leftrightarrow\left|2x-6\right|=10-5x\)

\(\left|2x-6\right|=\left[{}\begin{matrix}2x-6\left(đk:x\ge3\right)\\6-2x\left(đk:x< 3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=10-5x\left(đk:x\ge3\right)\\6-2x=10-5x\left(đk:x< 3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{7}\left(đk:x\ge3\right)\\x=\dfrac{4}{3}\left(đk:x< 3\right)\end{matrix}\right.\)

Th1 hiển nhiên hoại theo đk

x=4/3