\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

câu 2 có nghiệm x=2 , liên hợp đi 

câu 2 đề sai

ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !

câu 1 ) thì đúng

câu 2 sai đề

bài 1 chắc bạn sai đề. Mình lười lắm nên link đây nhé https://diendantoanhoc.net/topic/96618-sqrtx8frac3x27x84x2/

\(4x^2-4-3x=\sqrt[3]{x^2\left(x^2-1\right)}\)

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)-3x=\sqrt[3]{x^2\left(x-1\right)\left(x+1\right)}\)

dat \(\left(x-1\right)\left(x+1\right)=y\)

\(4y-3x=\sqrt[3]{x^2y}\)

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

\(\Leftrightarrow64y^3-144y^2x+108yx^2-27x^3=x^2y\)

\(\Leftrightarrow64y^3-144y^2x+107yx^2-27x^3=0\)

\(\Leftrightarrow64y^3-64y^2x-80y^2x+80x^2y+27x^2y-27x^3=0\)

\(\Leftrightarrow\left(y-x\right)\left(64y^2-80xy+27x^2\right)=0\)

de thay \(64y^2-80xy+27x^2=\left(8y\right)^2-2.8y.5x+25x^2+2x^2=\left(8y-5x\right)^2+2x^2>0\)

\(\Rightarrow y=x\)hay \(\left(x-1\right)\left(x+1\right)=x\Rightarrow x^2-x-1=0\) 

\(\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+1}{2}\\x=\frac{-\sqrt{5}+1}{2}\end{cases}}\)

câu b tương tự nhé bạn

ĐKXĐ: x lớn hơn hoặc bằng -1 và x nhỏ hơn hoặc bằng 1.

\(4+2\sqrt{1-x}=-3x+5\sqrt{x+1}+\sqrt{1-x^2}\)

\(\Leftrightarrow4+2\left(\sqrt{1-x}-1\right)+2=-3x+5\left(\sqrt{x+1}-1\right)+\left(\sqrt{1-x^2}-1\right)+5+1\)

\(\frac{-2x}{\sqrt{1-x}+1}=-3x+\frac{5x}{\sqrt{x+1}+1}-\frac{x^2}{\sqrt{1-x^2}+1}\Leftrightarrow x\left(\frac{x}{\sqrt{1-x^2}+1}-\frac{5}{\sqrt{x+1}+1}-\frac{2}{\sqrt{1-x}+1}+3\right)=0\)

\(\Leftrightarrow x=0.\)

\(Pt\Leftrightarrow3\left(x+1\right)+2\sqrt{1-x}+1=5\sqrt{x+1}+\sqrt{1-x^2}\)

đặt \(\sqrt{x+1}=a,\sqrt{1-x}=b\)

\(\Leftrightarrow3a^2+2b+1=a\left(5+b\right)\)

\(\Leftrightarrow3a^2-\left(5+b\right)a+2b+1=0\)

\(\Delta=b^2-4ac=\left(-b-5\right)^2-4.3.\left(2b+1\right)\)

\(=b^2+10b+25-24b-12\)

\(=b^2-14b+13\)

\(TH1:\Rightarrow a=\frac{5+b+\sqrt{b^2-14b+13}}{6}\)

\(\Rightarrow6a-5-b=\sqrt{b^2-14b+13}\)

\(\Rightarrow6\sqrt{1+x}-5-\sqrt{1-x}=\sqrt{1-x-14\sqrt{1-x}+13}\)

\(\hept{\begin{cases}x=0\left(nhan\right)\\x=......\left(loai\right)\end{cases}}\)

TH2:\(a=\frac{5+b-\sqrt{b^2-14b+13}}{6}\)

\(.............................................\)

cách này hơi dài.

Ta có \(\left(a+b+c+1\right)^2=\left(\left(a+1\right)+b+c\right)^2\)

\(=a^2+2a+1+b^2+c^2+2b+2c+2ab+2bc+2ac\left(f\right)\)

Từ \(\left(f\right)\Rightarrow2a+2b+2c+2ab+2bc+2ac\ge3\left(a^2+b^2+c^2\right)\)

Mà \(a,b,c\in\left[0;1\right]\)nên \(a\ge a^2,b\ge b^2,c\ge c^2\left(g\right)\)

Từ \(\left(f\right)vs\left(g\right)\Rightarrow2ab+2bc+2ac\ge a^2+b^2+c^2\)

Áp dụng bất đẳng thức cô si ta có:

\(2ab+2bc+2ac\ge3\sqrt[3]{8\left(abc\right)^2}=6\sqrt[3]{\left(abc\right)^2}\)

\(a^2+b^2+c^2\ge3\sqrt[3]{\left(abc\right)^2}\)

\(\Rightarrow2ab+2bc+2ac\ge a^2+b^2+c^2\Rightarrowđpcm\).Dấu bằng tự tìm nha.