\(B=\frac{2}{x-1}.\sqrt{\frac{x^2-2x+1}{4x^2}}\)

\(=\frac{2}{x-1}.\sqrt{\frac{\left(x-1\right)^2}{4x^2}}=\frac{2}{x-1}.\frac{1-x}{2x}=-\frac{1}{x}\)

\(P=-2\left[\left(1-x\right)-\sqrt{1-x}+\frac{1}{4}\right]+2+\frac{1}{2}=-2\left(\sqrt{1-x}-\frac{1}{2}\right)^2+\frac{5}{2}\le\frac{5}{2}\)

Max P = 5/2 khi 1-x =1/4 =>x =3/4

\(\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(x-1\right)^2}{81}}=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{2\sqrt{m}.\left|x-1\right|}{9}=\frac{2m}{9}\)

\(x\ne1\) chứ không phải x>1 nên không thể ghi |x-1|=x-1 nhé Despacito

A..mk vua nghi ra bai nay

\(\sqrt{\frac{m}{x^2-2x+1}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{\sqrt{4m\left(x-1\right)^2}}{9}\) ( Thoa man DKXD \(m>0;x\ne1\)

\(=\frac{\sqrt{m}}{x-1}.\frac{2\left(x-1\right).\sqrt{m}}{9}\)

\(=\frac{2m}{9}\)

ko biet co dung ko nua 

Mình làm cách khác mà cũng ra giống kq zậy đó

Mấy bạn giải chi tiết dùm mình , mình cần gấp lắm ạk

Cho \(5\sqrt{x}7\) mk viet nham

Sua lai thanh \(5\sqrt{x}-7\)

a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)

=>10 căn x+5-5 chia hết cho 2 căn x+1

=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)

hay \(x\in\varnothing\)

a.

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\frac{4\pi}{3};\frac{5\pi}{3}\right\}\)

b.

\(\Leftrightarrow sin2x=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-30^0+k360^0\\2x=210^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-15^0+k180^0\\x=105^0+k180^0\end{matrix}\right.\)

Pt vô nghiệm trên khoảng đã cho

c.

\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{0;\frac{2\pi}{3}\right\}\)

d.

\(\Leftrightarrow cos^2x\left(cosx-2\right)=0\)

\(\Leftrightarrow cosx=0\)

\(\Leftrightarrow x=90^0+k180^0\)

\(\Rightarrow x=\left\{90^0;270^0;450^0;630^0\right\}\)

Với x < 0, có:

\(2x^2\sqrt{\frac{9}{x^4}}\) = \(\frac{2x^2.9}{x^2}=18\)

Có \(2x^2+5x+3=2x^2+2x+3x+3=2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(2x+3\right)\)
\(\Rightarrow\left(\sqrt{2x+3}-\sqrt{x+1}\right)\left(\sqrt{2x^2+5x+3}+1\right)=x+2\left(ĐKXĐ:x\ge-1\right)\\ \Leftrightarrow\left(\sqrt{2x+3}-\sqrt{x+1}\right)\left(\sqrt{\left(2x+3\right)\left(x+1\right)}+1\right)=2x+3-\left(x+1\right)\left(1\right)\)

Đặt \(\sqrt{2x+3}=a\ge1,\sqrt{x+1}=b\ge0\), phương trình (1) trở thành:
\(\left(a-b\right)\left(ab+1\right)=a^2-b^2\)
\(\left(a-b\right)\left(ab+1\right)-\left(a-b\right)\left(a+b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left[a\left(b-1\right)-\left(b-1\right)\right]=0\\ \Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
+) Với a=b ta có: \(\sqrt{2x+3}=\sqrt{x+1}\Leftrightarrow2x+3=x+1\Leftrightarrow x=-2\left(ktm\right)\)
+) Với a=1 ta có: \(\sqrt{2x+3}=1\Leftrightarrow2x+3=1\Leftrightarrow x=-1\left(tm\right)\)
+) Với b=1 ta có : \(\sqrt{x+1}=1\Leftrightarrow x+1=1\Leftrightarrow x=0\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-1;0\right\}\).
Tick cho mình nha <3 !!!