Tìm x: \(5x^2+7,1=\sqrt{49}\)
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Tìm x:
a) 15 - 4x2 = - 21
b) 5x2 + 7,1 = \(\sqrt{49}\)
c) 4,7 + 3\(\sqrt{x}\) = 5,9
d) 2,8 - 1,5\(\sqrt{x}\) = \(\sqrt{\frac{98}{8}}\)
a, 4x2=15-(-21)
=36
x2=36:4
x2=4
x2=22
x=2
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b. 5x2+7,1=\(\sqrt{49}\)
\(\Rightarrow\)5x2+7,1=7
\(\Rightarrow\)5x2 = 7+7,1
\(\Rightarrow\)5x2 =14,1
\(\Rightarrow\)x2 =\(\dfrac{14,1}{5}\)
\(\Rightarrow\)x =\(\sqrt{\dfrac{14,1}{5}}\)
cho mk 1 tick đúng và câu tiếp thao sẽ hiện ra
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a) 129 - 50x^2 = -159
b)5x^2 + 7,1 =\(\sqrt{49}\)
c) 4,7 + 3\(\sqrt{x}\)= 5,9
d) 2,8 - 1,5\(\sqrt{x}\)=\(\sqrt{\frac{98}{8}}\)
Tìm điều kiện có nghĩa:1) sqrt{2x^2}2) sqrt{-x}3) sqrt{-x^2-3}4) sqrt{x^2+2x+3}5) sqrt{-a^2+8a-16}6) sqrt[]{16x^2-25}7) sqrt{4x^2-49}8) sqrt{8-x^2}9) sqrt{x^2-12}10) sqrt{x^2+2x-3}11) sqrt{2x^2+5x+3}12) sqrt{dfrac{4}{x-1}}13) sqrt{dfrac{-1}{x-3}}14) sqrt{dfrac{-3}{x+2}}15) sqrt{dfrac{1}{2a-1}}16) sqrt{dfrac{2}{3-2a}}17) sqrt{dfrac{-1}{2a-5}}18) sqrt{dfrac{-2}{3-5a}}19) sqrt{dfrac{-a}{5}}20) dfrac{1}{sqrt{-3a}}
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Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
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tìm giá trị lớn nhất . biết x>=49
\(\frac{\sqrt{x-49}}{5x}\)
Tìm x ,biết:
a)\(5x\):\(\frac{3}{5}=\frac{4}{7}:\sqrt{\frac{4}{49}}\)
b)\(0,15:3\sqrt{x}=0,3:\frac{2}{3}\)
a,\(\frac{25x}{3}=\frac{\frac{4}{7}}{\frac{2}{7}}=7\)=> x=\(\frac{21}{25}\)
b,\(\frac{0,15}{3\sqrt{x}}=\frac{0,3\cdot3}{2}\)<=>\(0,3=0,3\cdot9\sqrt{x}\)Hay \(\sqrt{x}=\frac{1}{9}=>x=\frac{1}{3}\)
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Giải PT:
a) -5x+7\(\sqrt{x}\) +12=0
b) \(\dfrac{1}{3}\)\(\sqrt{4x^2-20}\) +2\(\sqrt{\dfrac{x^2-5}{9}}\) -3\(\sqrt{x^2-5}=0\)
c) \(\sqrt{9x+27}+5\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=5\)
d) \(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=3\sqrt{x-2}+8\)
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
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d. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 2\sqrt{x-2}=8$
$\Leftrightarrow \sqrt{x-2}=4$
$\Leftrightarrow x=4^2+2=18$ (tm)
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b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
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Xem thêm câu trả lời
Tìm x:
25x^2 - 49 = (5x - 7)(x + 3)
\(\Leftrightarrow\left(5x-7\right)\left(5x+7-x-3\right)=0\)
\(\Leftrightarrow\left(5x-7\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)
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\(25x^2-49=\left(5x-7\right)\left(x+3\right)\)
\(\Rightarrow\left(5x\right)^2-7^2=\left(5x-7\right)\left(x+3\right)\)
\(\Rightarrow\left(5x-7\right)\left(5x+7\right)-\left(5x-7\right)\left(x+3\right)=0\)
\(\Rightarrow\left(5x-7\right)\left(5x+7-x-3\right)=0\)
\(\Rightarrow\left(5x-7\right)\left(4x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-7=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=7\\4x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{7}{5};-1\right\}\)
tích cho mk nhé
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Tìm x, biết:
a)\(\sqrt{4-5x}=12\)
b)\(\sqrt{x^2-14x+49}\)\(-3x=1\)
c)\(\sqrt{9x-9}\)+\(2\sqrt{16x-16}\)\(-\frac{1}{3}\sqrt{25x-25}\)\(=\sqrt{x-1}\)
a, \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=140\Leftrightarrow x=28\)
b,ĐK : \(x\ge7\)
\(\sqrt{x^2-14x+49}-3x=1\Leftrightarrow\sqrt{\left(x-7\right)^2}=3x+1\)
\(\Leftrightarrow x-7=3x+1\Leftrightarrow-2x-8=0\Leftrightarrow x=-4\)( vô lí )
c, Bn làm nốt nhé
a) đk: \(x\le\frac{4}{5}\)
Ta có: \(\sqrt{4-5x}=12\)
\(\Leftrightarrow\left|4-5x\right|=144\)
\(\Rightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
\(\Rightarrow x=-28\left(tm\right)\)
b) Ta có: \(\sqrt{x^2-14x+49}-3x=1\)
\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=1+3x\)
\(\Leftrightarrow\left|x-7\right|=3x+1\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=3x+1\\x-7=-3x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-8\\4x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)
CTRL là cái gì vậy
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Tìm x, biết:a) left(5x+1right)^2dfrac{36}{49}b) left[left(-0,5right)^3right]^xdfrac{1}{64}c) 2020^{left(x-2right).left(2x+3right)}1d) left(x+1right)^{x+10}left(x+1right)^{x+4} với xin Ze) dfrac{3}{4}sqrt{x}-dfrac{1}{2}dfrac{1}{3}
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Tìm x, biết:
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)
d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in Z\)
e) \(\dfrac{3}{4}\sqrt{x}-\dfrac{1}{2}=\dfrac{1}{3}\)
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
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