\(\left(2x+3\right)^2=\frac{169}{121}\)
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\(\left(2x+3\right)^2=\frac{9}{121}\\ \left(3x-1\right)^3=-\frac{8}{27}\)
a)\(\left(2x+3\right)^2=\frac{9}{121}\\ \Leftrightarrow\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{matrix}\right.\)
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b)\(\left(3x-1\right)^3=\frac{-8}{27}\\ \Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\\ 3x-1=\frac{-2}{3}\\ \Rightarrow x=\frac{1}{9}\)
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a) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Rightarrow2x+3=\pm\frac{3}{11}\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{3}{11}-3=-\frac{30}{11}\\2x=\left(-\frac{3}{11}\right)-3=-\frac{36}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-\frac{30}{11}\right):2\\x=\left(-\frac{36}{11}\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}.\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(\Rightarrow3x=\left(-\frac{2}{3}\right)+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}.\)
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Thực hiện phép tính sau
a. F=[12(1)-2,3(6)]:4,(21)
b.\(\frac{1\frac{11}{34}.4\frac{3}{7}-\left(\frac{3}{2}-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}.\left(12-5\frac{1}{3}\right)}\)
c.1-\(\frac{\sqrt{121}}{\sqrt{196}}-\frac{\sqrt{169}}{\sqrt{144}}+\frac{\sqrt{25}}{\sqrt{36}}+\left(-1\frac{2}{3}\right):\left(-3\frac{1}{3}\right)\)
c/
\(=1-\frac{11}{14}-\frac{14}{12}+\frac{5}{6}+\frac{-5}{3}:\frac{-10}{3}\)
\(=1-\frac{11}{14}-\frac{14}{12}+\frac{5}{6}+\frac{-5}{3}.\frac{-3}{10}\)
\(=1-\frac{11}{14}-\frac{14}{12}+\frac{5}{6}+\frac{1}{2}\)
\(=1-\left(\frac{66}{84}+\frac{98}{84}-\frac{70}{84}-\frac{42}{84}\right)\)
Mik làm tiếp nhé tại lúc nãy bấm nhầm!
Câu c/ (tiếp theo)
\(=1-\frac{52}{84}\)
\(=\frac{84}{84}-\frac{52}{84}=\frac{32}{84}=\frac{8}{21}\)
Câu a: Sai đề
Cho \(\frac{a}{x+y}=\frac{13}{x+z}\) và \(\frac{169}{\left(x+z\right)^2}=\frac{-27}{\left(z-y\right)\left(2x+y+z\right)}\) . Tính \(A=\frac{2a^3-12a^2+17a-2}{a-2}\)
1A)thực hiện phép tính
a)\(\sqrt{144}.\sqrt{-\frac{-49}{64}}.\sqrt{0,01}\)
b)\(\left(\sqrt{0,25}-\sqrt{\left(-15\right)^2}+\sqrt{2,25}\right):\sqrt{169}\)
1b)hãy tính
a)\(\left(\sqrt{0,04}-\sqrt{\left(-1,2\right)^2}+\sqrt{121}\right).\sqrt{81}\)
b)\(75:\sqrt{3^2+\left(-4\right)^2}-3.\sqrt{\left(-5\right)^2-3^2}\)
Cho \(\frac{a}{x+y}=\frac{13}{x+z}\) và \(\frac{169}{\left(x+z\right)^2}=\frac{-27}{\left(z-y\right)\left(2x+y+z\right)}\) . Tính \(A=\frac{2a^3-12a^2+17a-2}{a-2}\)
Cho \(\frac{a}{x+y}=\frac{13}{x+z}\)và \(\frac{169}{\left(x+z\right)^2}=\frac{-27}{\left(z-y\right)\left(2x+y+z\right)}\)
Tính giá trị của biểu thức: \(E=\frac{2a^3-12a^2+17a-2}{a-2}\)
Tìm số nguyên N biết:
\(a,\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(b,\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(c,\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
\(d,\left(2x+3\right)^2=\frac{9}{121}^n\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
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tìm x:
a) \(\left(2x+3\right)^2=\frac{9}{121}\)
b)\(\left(3x-1\right)^3=-\frac{8}{27}\)
a, \(\left(2x+3\right)^2=\frac{3^2}{11^2}\)
từ đó suy ra
\(2x+3=\frac{3}{11}\)
2x=3/11-3
2x=-2/8/11
x=-2/8/11:2
x=-1/4/11
b,
(3x-1)^3=-8/27
(3x-1)^3=(-2/3)^3
Vậy suy ra
3x-1=-2/3
3x=-2/3+1
3x=1/3
x=1/3:3
x=1/9
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1, \(\left(2x-3\right)^4=\left(2x-3\right)^6\)
2 , \(\frac{\left(-2\right)^n}{16}=32\)
3 , \(\frac{8}{25}=\frac{2x}{5^{x-1}}\)
4 , \(\frac{64}{169}=\left(\frac{-8}{13}\right)^x\)
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