\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

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

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

ĐKXĐ: \(x\ge2\)

Đặt \(u=\sqrt{x+3};v=\sqrt{x-2}\) Phương trình trở thành :

\(\left(u-v\right)\left(1+uv\right)=5\) Mặt khác ta thấy \(u^2-v^2=5\)

\(\Rightarrow\left(u-v\right)\left(1+uv\right)=\left(u-v\right)\left(u+v\right)\) (*)

vì \(u-v>0\) nên chia cả hai vế (*) cho \(u-v\)

Ta được: \(1+uv=u+v\) \(\Leftrightarrow uv-u-\left(v-1\right)=0\Leftrightarrow\left(v-1\right)\left(u-1\right)=0\)

\(\left[{}\begin{matrix}u=1\\v=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=1\\x-2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(Loai\right)\\x=3\end{matrix}\right.\)

Vậy phương trình có nghiệm duy nhất \(x=3\)

câu a tớ giải được rồi, các bn giải câu b giùm mk

@Rain Tờ Rym Te

Đề sai

Đặt \(u=\sqrt{x+1};t=\sqrt{1-x};\text{đ}k:-1\le x\le1\)

Phương trình trở thành:

\(u+2u^2=-t^2+t+3ut\Leftrightarrow\left(u-t\right)^2+u\left(u-t\right)+\left(u-t\right)=0\)

\(\Leftrightarrow\left(u-t\right)\left(2u-t+1\right)=0\Leftrightarrow\orbr{\begin{cases}u=t\\2u+1=t\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-24}{25}\end{cases}}}\)

mình dùng cách khác nhé :((

\(\sqrt{x+1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\left(đk:-1\le x\le1\right)\)

\(< =>\sqrt{x+1}-1+2x+2-3=x-1+\sqrt{1-x}-1+3\sqrt{1-x^2}-3\)

\(< =>\frac{x}{\sqrt{x+1}+1}+2x-1-x+1=-\frac{x}{\sqrt{1-x}+1}+\frac{9\left(1-x^2-1\right)}{3\sqrt{1-x^2}+3}\)

\(< =>\frac{x}{\sqrt{x+1}+1}+x+\frac{x}{\sqrt{1-x}+1}+\frac{9x^2}{3\sqrt{1-x^2}+3}=0\)

\(< =>x\left(\frac{1}{\sqrt{x+1}+1}+1+\frac{1}{\sqrt{1+x}+1}+\frac{9x}{3\sqrt{1-x^2}+3}\right)=0< =>x=0\)

rồi đến đây dùng đk đánh giá cái ngoặc khác 0 là ok