Tìm x :
\(x+2=3\sqrt{1-x^2}+\sqrt{1+x}\)
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Những câu hỏi liên quan
28 tháng 8 2023
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
H9
HT.Phong (9A5)
CTVHS
21 tháng 9 2023
\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0\)
\(\Leftrightarrow\dfrac{5}{5\sqrt{x}+10}-\dfrac{\sqrt{x}+2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{5-\sqrt{x}-2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-3\right)}{5\sqrt{x}+10}>0\)
Mà: \(5\sqrt{x}+10\ge10>0\forall x\)
\(\Leftrightarrow\sqrt{x}>3\)
\(\Leftrightarrow x>9\)
_________
\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{4}{2\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)}{2\sqrt{x}+6}< 0\)
Mà: \(2\sqrt{x}+6\ge6>0\forall x\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
\(\Leftrightarrow0\le x\le1\)
15 tháng 1 2022
a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)
b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)
=>x<1
23 tháng 9 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
b: Ta có: \(P=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
1 tháng 10 2021
1: ĐKXĐ: \(-1< x< 1\)
2: ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)
3: ĐKXĐ: \(\left[{}\begin{matrix}x< -3\\x\ge2\end{matrix}\right.\)
4: ĐKXĐ: \(2< a\le3\)
MH
25 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)
\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)
\(\left(\sqrt{x-1}+\sqrt{x+3}+3\right)\left(\sqrt{x-1}+\sqrt{x+3}-2\right)=0\)
Đến đây em xét các trường hợp rồi bình phương lên là được nha
b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)
\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)
\(\left(\sqrt{3x-2}+\sqrt{x-1}-3\right)\left(\sqrt{3x-2}+\sqrt{x-1}+2\right)=0\)
Đến đây em xét các trường hợp rồi bình phương lên là được nha
AH
Akai Haruma
Giáo viên
25 tháng 9 2023
a/ ĐKXĐ: $x\geq 1$
Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:
$a+b+2ab=6-(a^2+b^2)$
$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$
$\Leftrightarrow (a+b)^2+(a+b)-6=0$
$\Leftrightarrow (a+b-2)(a+b+3)=0$
Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$
$\Leftrightarrow a+b=2$
Mà $b^2-a^2=(x+3)-(x-1)=4$
$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$
$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$
$\Leftrightarrow x=1$ (tm)
1 tháng 8 2021
1) Ta có: \(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}-1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+2\sqrt{x}-3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}-2\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-2}{\sqrt{x}-3}\)
Sửa đề: \(x=7+4\sqrt{3}\)
Thay \(x=7+4\sqrt{3}\) vào P, ta được:
\(P=\dfrac{3\left(2+\sqrt{3}\right)-2}{2+\sqrt{3}-3}=\dfrac{6+3\sqrt{3}-2}{\sqrt{3}-1}\)
\(=\dfrac{4+3\sqrt{3}}{\sqrt{3}-1}=\dfrac{13+7\sqrt{3}}{2}\)
24 tháng 8 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
31 tháng 10 2021
\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)
31 tháng 10 2021
a: \(=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
25 tháng 8 2021
a: Ta có: \(2\sqrt{2}-\dfrac{1}{2}\cdot\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\cdot\dfrac{1}{2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x}=4\sqrt{2}\)
hay x=32
b: Ta có: \(2\sqrt{x}-\sqrt{\dfrac{x}{3}}=1\)
\(\Leftrightarrow2\sqrt{x}-\dfrac{\sqrt{3}}{3}\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{6+\sqrt{3}}{11}\)
hay \(x=\dfrac{39+12\sqrt{3}}{121}\)
c: Ta có: \(4\sqrt{x}+\sqrt{\dfrac{x}{2}}=\dfrac{1}{3}\)
\(\Leftrightarrow4\sqrt{x}+\dfrac{\sqrt{2}}{2}\sqrt{x}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}=\dfrac{8-\sqrt{2}}{93}\)
hay \(x=\dfrac{66-16\sqrt{2}}{8649}\)
\(x+2=3\sqrt{1-x^2}+\sqrt{1+x}\)
\(ĐKXĐ:-1\le x\le1\)
\(x+2=3\sqrt{1-x}\sqrt{1+x}+\sqrt{1+x}\)
\(\left(3\sqrt{1-x}\sqrt{1+x}-\frac{3}{2}\right)+\left(\sqrt{1+x}-x-\frac{1}{2}\right)=0\)
\(\frac{9\left(1-x\right)\left(1+x\right)-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-\left(x+\frac{1}{2}\right)^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{9-9x^2-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-x^2-x-\frac{1}{4}}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{\frac{27}{4}-9x^2}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{9\left(\frac{3}{4}-x^2\right)}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\left(\frac{3}{4}-x^2\right)\left(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}\right)=0\)
\(\orbr{\begin{cases}\frac{3}{4}-x^2=0\\\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}=0\left(KTM\right)\end{cases}< =>x=\frac{\sqrt{3}}{2}\left(TM\right)}\)
\(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}>0\)nên pt ktm