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a) Ta có: \(\left(x-\sqrt{2}\right)+3\left(x^2-2\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\3x+3\sqrt{2}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\3x=-3\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=\dfrac{-3\sqrt{2}-1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)

b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

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

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

\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{5}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)

24 tháng 2 2019

Đây là toán 9 chứ

NV
16 tháng 4 2022

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

NV
16 tháng 4 2022

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

NV
28 tháng 2 2021

Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:

\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)

\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)

Ta có:

\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)

\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)

Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)

Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)

10 tháng 11 2017

\(x^2-2-2\sqrt{4x-7}=0\)

\(\Leftrightarrow\left(4x-7-2\sqrt{4x-7}+1\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(\sqrt{4x-7}-1\right)^2+\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{4x-7}-1=0\\x-2=0\end{matrix}\right.\)

Tự làm tiếp nhé.

. . .

\(4x^2-5x+1+2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)+2\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left[\left(4x-1\right)\sqrt{x-1}+2\right]=0\)

\(\Rightarrow x=1\)

. . .

\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)

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

\(\Leftrightarrow\left|x-2\right|+\left|x-3\right|=1\)

\(VT=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1=VP\)

Dấu "=" xảy ra khi \(\left(x-2\right)\left(3-x\right)\ge0\)

Đến đây lập bảng xét dấu

. . .

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

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

Tự làm tiếp nhé.

10 tháng 11 2017

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14-5\right)=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)\left(x-5\right)=0\)

\(\Rightarrow x=5\)

. . .

\(\sqrt{2x^2-4x+5}-x+4=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\2x^2-4x+5=x^2-8x+16\end{matrix}\right.\)

Tự làm tiếp nhé.

. . .

\(\sqrt{2x+3}+\sqrt{x-1}=\sqrt{x+6}\)

\(\Leftrightarrow\sqrt{2x+3}=\sqrt{x+6}-\sqrt{x-1}\)

\(\Leftrightarrow2x+3=x+6-2\sqrt{\left(x+6\right)\left(x-1\right)}+x-1\)

\(\Leftrightarrow2\sqrt{x^2+5x-6}=2\)

\(\Leftrightarrow x^2+5x-6=1\)

Tự làm tiếp nhé.

. . .

\(x+y+\dfrac{1}{2}=\sqrt{x}+\sqrt{y}\)

\(\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\left(y-\sqrt{y}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\left(\sqrt{y}-\dfrac{1}{2}\right)^2=0\)

Tự làm tiếp nhé.

22 tháng 9 2020

\(BPT\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(3x^2+2x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)

\(\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(x+1\right)\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)

\(\Leftrightarrow\left(x+1\right)\text{[}2+\sqrt{x^2-2x+5}+\frac{2x\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\text{]}\le0\)

\(\Leftrightarrow\left(x+1\right)\left(4\sqrt{x^2+1}+2\sqrt{x^2-2x+5}+2\sqrt{\left(x^2+1\right)\left(x^2-2x+5\right)}+7x^2-4x+5\right)\)\(\le0\Leftrightarrow x+1\le0\Leftrightarrow x\le-1\)

11 tháng 12 2017

nhầm,\(\sqrt{4x^2+9x+2}\)

24 tháng 4 2020

ĐK : | x| \(\ge\sqrt{7}\)

x + 4x - 7 = ( x + 4 ) \(\sqrt{x^2-7}\)

\(\Leftrightarrow\left(x^2-7\right)+4x-\left(x+4\right)\sqrt{x^2-7}=0\)

\(\Leftrightarrow\left(x^2-7\right)+4x-x\sqrt{x^2-7}-4\sqrt{x^2-7}=0\)

\(\Leftrightarrow\sqrt{x^2-7}\left(\sqrt{x^2-7}-x\right)-4\left(\sqrt{x^2-7}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2-7}-x\right)\left(\sqrt{x^2-7}-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-7}-x=0\\\sqrt{x^2-7}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-7}=x\\\sqrt{x^2-7}=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2-7=x^2\\x^2-7=16\end{cases}}}\)

<=> x2 =23 <=> x = \(\pm\sqrt{23}\)( T/m đk)

26 tháng 4 2020

Có thể đặt \(t=\sqrt{x^2-7}\left(t\ge0\right)\)cho dễ nhìn