\(\sqrt{x+1}+2\)=0
\(3\sqrt{x+1}=40\)
Những câu hỏi liên quan
1, \(x^3-x-3=2\sqrt{6x-x^2}\)
2, \(x^3+6x^2-171x-40\left(x+1\right)\sqrt{5x-1}+20=0\)
3, \(\sqrt[3]{x+3}+\sqrt[3]{x-3}=\sqrt[5]{x-5}+\sqrt[5]{x+5}\)
4. \(\left(\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{x+1}\right)^2=\frac{4\left(1+\sqrt{1+4x}\right)}{x+1+\sqrt{x^2+3x+2}}\)
GIẢI PHƯƠNG TRÌNH
a) \(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\)
b) \(\sqrt{9x^2+12x+4}=4x\)
c) \(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
d) \(\sqrt{5x-6}-3=0\)
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
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\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
\(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\left(tmdk\right)\)
b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x\)
\(\Leftrightarrow\left|3x-2\right|=3x\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=3x\\3x-2=-3x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)
Các câu còn lại làm tương tự nhé
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\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
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\(2x^4+8x=4\sqrt{4+x^4}+4\sqrt{x^4-4}\)
\(^{x^3-3x^2-8x+40-8\sqrt[4]{4x+4}=0}\)
\(\sqrt[4]{x}+\sqrt[4]{1-x}+\sqrt{x}-\sqrt{1-x}=\sqrt{2}+\sqrt[4]{8}\)
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
Giải phương trình:
1) \(x^2-4x-2\sqrt{2x-5}+5=0\)
2)\(x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
3)\(\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
4)\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
\(1.x^2-4x-2\sqrt{2x-5}+5=0\left(x>=\dfrac{5}{2}\right)\)
\(\text{⇔}2x-5-2\sqrt{2x-5}+1+x^2-6x+9=0\)
\(\text{⇔}\left(\sqrt{2x-5}-1\right)^2+\left(x-3\right)^2=0\)
\(\text{⇔}\sqrt{2x-5}-1=0\) hoặc \(x-3=0\)
\(\text{⇔}x=3\left(TM\right)\)
KL...........
\(2.x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
\(\text{⇔}x-2\sqrt{x}+1+y-1-4\sqrt{y-1}+4=0\)
\(\text{⇔}\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2=0\)
\(\text{⇔}x=1;y=5\)
KL..........
\(3.\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
\(\text{⇔}2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-5}=x+y+z-7\)
\(\text{⇔}x-2-2\sqrt{x-2}+1+y-3-2\sqrt{y-3}+1+z-5-2\sqrt{z-5}+1=0\)
\(\text{⇔}\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-5}-1\right)^2=0\)
\(\text{⇔}x=1;y=4;z=6\)
KL...........
\(d.Tuong-tự-nhé-bn\)
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Tìm x biết :
\(3x\sqrt{x+1}=40\)
\(\sqrt{x+1}+2=0\)
\(\sqrt{\left(x+1\right)^2}=3\)
\(\sqrt{x-3}=4\)
b: =>căn x+1=-2(loại)
c: =>|x+1|=3
=>x+1=3 hoặc x+1=-3
=>x=-4 hoặc x=2
d: =>x-3=16
=>x=19
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rút gọn các biểu thức
\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)(vs x>= 0;y>=0;x khác y)
\(\frac{2}{2a-1}.\sqrt{5a^2\left(1-4a+4a^2\right)}\)(a>0,5)
1, \(\sqrt{x-1}+\sqrt{x-4}=5\)
2, \(2x-7\sqrt{x}+5=0\)
3, \(\sqrt{2x+1}+\sqrt{x-3}=2\sqrt{x}\)
4, \(x-4\sqrt{x}+2021\sqrt{x-4}+4=0\)
5, \(\sqrt{2x-3}-\sqrt{x+1}=7\left(4-x\right)\)
1. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$
$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$
$\Leftrightarrow 22=10\sqrt{x-4}$
$\Leftrightarrow 2,2=\sqrt{x-4}$
$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$
(thỏa mãn)
2. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$
$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$
$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$
$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)
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3. ĐKXĐ: $x\geq 3$
Bình phương 2 vế thu được:
$3x-2+2\sqrt{(2x+1)(x-3)}=4x$
$\Leftrightarrow 2\sqrt{(2x+1)(x-3)}=x+2$
$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$
$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$
$\Leftrightarrow (x-4)(7x+4)=0$
Do $x\geq 3$ nên $x=4$
Thử lại thấy thỏa mãn
Vậy $x=4$
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4. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$
$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$
Ta thấy, với mọi $x\geq 4$ thì:
$(\sqrt{x}-2)^2\ge 0$
$2021\sqrt{x-4}\geq 0$
Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
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Xem thêm câu trả lời
Bài 1: Rút gọna. left(5-2sqrt{3}right)^2+left(5+2sqrt{3}right)^2b. left(sqrt{5}+sqrt{2}right)^2-left(2sqrt{5}+1right)left(2sqrt{5}-1right)-sqrt{40}c. left(sqrt{2}-1right)^2-frac{2}{3}sqrt{4}+frac{4sqrt{2}}{5}+sqrt{1frac{11}{15}}-sqrt{2}d. left(sqrt{6}-sqrt{18}+5sqrt{2}-frac{1}{2}sqrt{8}right)2sqrt{6}+2sqrt{3}e. left(2sqrt{3}-3sqrt{2}right)^2+6sqrt{6}+3sqrt{24}Bài 2: Rút gọnA left(frac{1}{x-sqrt{x}}+frac{1}{sqrt{x}-1}:frac{sqrt{x+1}}{x-2sqrt{x}+1}right)(x0 ; x khác 1)
Đọc tiếp
Bài 1: Rút gọn
a. \(\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2\)
b. \(\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}\)
c. \(\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}\)
d. \(\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}\)
e. \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}\)
Bài 2: Rút gọn
A =\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{\sqrt{x+1}}{x-2\sqrt{x}+1}\right)\)(x>0 ; x khác 1)