\(1.\sqrt{16-8x+x^2}=4-x\)

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

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

\(\)\(\sqrt{\left(2x-3\right)^2}=2x-3\)

\(2x-3-2x+3=0\)

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)

a/ \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\sqrt{\left(4-x\right)^2}=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\left|4-x\right|=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le4\\\left[{}\begin{matrix}4-x=4-x\left(loại\right)\\4-x=x-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

Vậy...

b/ \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\sqrt{\left(2x-3\right)^2}=2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}2x-3=2x-3\left(loại\right)\\2x-3=3-2x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy...

1: Ta có: \(\sqrt{4x^2-12x+9}=3-2x\)

\(\Leftrightarrow\left(2x-3\right)^2=\left(3-2x\right)^2\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(3-2x\right)^2=0\)

\(\Leftrightarrow\left[\left(2x-3\right)-\left(3-2x\right)\right]\left[\left(2x-3\right)+\left(3-2x\right)\right]=0\)

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

\(\Leftrightarrow\left(4x-6\right)\cdot0=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

2) Ta có: \(\sqrt{x^2-2\cdot\sqrt{2}\cdot x+2}=\sqrt{9-4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}-\sqrt{1+2\cdot1\cdot\sqrt{2}+2}\)

\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\left|\sqrt{8}-1\right|-\left|1+\sqrt{2}\right|\)

\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8}-1-1-\sqrt{2}\)

\(\Leftrightarrow\left|x-\sqrt{2}\right|=\sqrt{2}-2\)(*)

Trường hợp 1: \(x\ge\sqrt{2}\)

(*)\(\Leftrightarrow x-\sqrt{2}=\sqrt{2}-2\)

\(\Leftrightarrow x-\sqrt{2}-\sqrt{2}+2=0\)

\(\Leftrightarrow x-2\sqrt{2}+2=0\)

\(\Leftrightarrow x=2\sqrt{2}-2\)(loại)

Trường hợp 2: \(x< \sqrt{2}\)

(*)\(\Leftrightarrow\sqrt{2}-x=\sqrt{2}-2\)

\(\Leftrightarrow\sqrt{2}-x-\sqrt{2}+2=0\)

\(\Leftrightarrow2-x=0\)

hay x=2(loại)

Vậy: S=∅

\(1.4x^2-12x+9=9-12x+4x^2\)

\(0x=0\)

Pt tm với mọi x

-9.75332792

minh moi hok lop 6 thôi ban ơi

ra số lẻ lắm

9,75332741792369

a.ĐKXĐ:\(\frac{3}{2}\le x\le\frac{5}{2}\)

AD BĐT Cauchy ta được:

\(\sqrt{\left(2x-3\right)1}\le\frac{2x-3+1}{2}=\frac{2x-2}{2}=x-1\)

\(\sqrt{\left(5-2x\right)\cdot1}\le\frac{5-2x+1}{2}=\frac{6-2x}{2}=3-x\)

Do đó \(\sqrt{2x-3}+\sqrt{5-2x}\le x-1+3-x=2\)(1)

Lại có \(3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)(2)

Từ (1) và (2) suy ra \(\sqrt{2x-3}+\sqrt{5-2x}\le3x^2-12x+14\)

Dấu = khi x=2 (tm ĐKXĐ)

PHẦN b giải tương tự

\(c,=2+2\sqrt{3}-\left(2+\sqrt{2}\right)=2\sqrt{3}-\sqrt{2}\\ d,=\sqrt{\left(2x-3\right)^2}-2x+1=\left|2x-3\right|-2x+1\\ =2x-3-2x+1=-2\left(x\ge\dfrac{3}{2}\Leftrightarrow2x-3\ge0\right)\)