(x-3)2=36/49
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5.x - 9 = 5 + 3.x
5x - 3x = 5 + 9
2x = 14
x = 14 : 2
x = 7
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(5x + 1)² = 36/49
5x + 1 = 6/7 hoặc 5x + 1 = -6/7
*) 5x + 1 = 6/7
5x = 6/7 - 1
5x = -1/7
x = -1/7 : 5
x = -1/35
*) 5x + 1 = -6/7
5x = -6/7 - 1
5x = -13/7
x = -13/7 : 5
x = -13/35
Vậy x = -13/35; x = -1/35
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2ˣ⁻¹ = 16
2ˣ⁻¹ = 2⁴
x - 1 = 4
x = 4 + 1
x = 5
\(\left[x-\frac{3}{2}\right]^3=\frac{1}{64}\)
\(\Leftrightarrow\left[x-\frac{3}{2}\right]^3=\left[\frac{1}{4}\right]^3\)
\(\Leftrightarrow x-\frac{3}{2}=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\left[x+\frac{3}{2}\right]^2=\frac{9}{49}\)
\(\Leftrightarrow\left[x+\frac{3}{2}\right]^2=\left[\frac{3}{7}\right]^2\)
\(\Leftrightarrow x+\frac{3}{2}=\pm\frac{3}{7}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{2}=\frac{3}{7}\\x+\frac{3}{2}=-\frac{3}{7}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{15}{14}\\x=-\frac{27}{14}\end{cases}}\)
Lời giải:
Gọi biểu thức là A.
\(A=256.\frac{1}{8}+\frac{1}{49^2}.7^3+\frac{1}{36^2}.\frac{1}{8^2}.27\\ =32+\frac{1}{7}+\frac{1}{3072}=32\frac{3079}{21504}\)
\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{3;-3\right\}\)
\(7,\left(3x+1\right)^2-16=0\\ \Leftrightarrow\left(3x+1\right)^2-4^2=0\\ \Leftrightarrow\left(3x+1-4\right)\left(3x+1+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)
\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-2;5\right\}\)
(5x+1)2 =\(\frac{36}{49}\)
(5x+1)2=\(\frac{6}{7}^2\)
=>(5x+1)=\(\frac{6}{7}\)
5x =\(\frac{6}{7}\)-1=\(\frac{-1}{7}\)
x = \(\frac{-1}{7}\) :5
x= \(\frac{-1}{35}\)
1. \(4x^2-49=0\)
\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\Leftrightarrow x=-\dfrac{7}{2}\\2x-7=0\Leftrightarrow x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)
===========
2. \(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x=6\)
Vậy: \(x=6\)
===========
3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)
\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)
1: Ta có: \(4x^2-49=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
2: Ta có: \(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x-6=0\)
hay x=6
đáp án là x=27/7
\(\left(x-3\right)^2=\frac{36}{49}\)
=> \(\left(x-3\right)^2=\left(\frac{6}{7}\right)^2\)
=> \(x-3=\frac{6}{7}\)
=> \(x=\frac{6}{7}+3\)
=> \(x=\frac{27}{7}\)