\(\left(2x-3\right)^2=36\)
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3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36
=> 3(x^2 + 4x + 4) + 4x^2 - 4x + 1 - 7(x^2 - 9) = 36
=> 3x^2 + 12x + 12 + 4x^2 - 4x + 1 - 7x^2 + 63 = 36
=> 8x + 76 = 36
=> 8x = -40
=> x = -5
\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
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 \(S=\left\{-2;5\right\}\)
A = (2x+3)(4x2−6x+9)−2(4x3−1)−36
=8x3-12x2+18x+12x2-18x+27-8x3+2-36
=-7
Ta có: \(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)-36\)
\(=8x^3+27-8x^3+2-36\)
\(=-7\)
a: \(P=-3x^3+5x\)
\(=x\cdot\left(-3x^2\right)+x\cdot5\)
\(=x\left(-3x^2+5\right)\)
b: \(Q=\left(2x-1\right)+\left(x-2\right)\left(2x-1\right)\)
\(=\left(2x-1\right)\left(1+x-2\right)\)
\(=\left(2x-1\right)\left(x-1\right)\)
c: \(R=4-16x^2\)
\(=4\cdot1-4\cdot4x^2\)
\(=4\left(1-4x^2\right)\)
\(=4\left(1-2x\right)\left(1+2x\right)\)
d: \(S=36-4x^2\)
\(=4\cdot9-4\cdot x^2\)
\(=4\left(9-x^2\right)\)
\(=4\left(3-x\right)\left(3+x\right)\)
e: \(T=8x^3-1\)
\(=\left(2x\right)^3-1^3\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right)\)
f: \(Q=8-x^3\)
\(=2^3-x^3\)
\(=\left(2-x\right)\left(4+2x+x^2\right)\)
g: \(N=64-x^3\)
\(=4^3-x^3\)
\(=\left(4-x\right)\left(16+4x+x^2\right)\)
1) (x - 2)2 - (x - 3)(x + 3) = 17
=> x2 - 4x + 4 - x2 + 9 = 17
=> -4x = 17 - 13
=> -4x = 4
=> x = -1
2) TTT
3) x2 + 6x - 147 = 0
=> x2 + 19x - 13x - 147 = 0
=> x(x + 19) - 13(x + 19) = 0
=> (x - 13)(x + 19) = 0
=> \(\orbr{\begin{cases}x-13=0\\x+19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=13\\x=-19\end{cases}}\)
4) (3x - 5)(2x + 3) - 6x2 = 7
=> 6x2 + 9x - 10x - 15 - 6x2 = 7
=> -x - 15 = 7
=> -x = 7 + 15
=> -x = 22
=> x = -22
5) TL
Vì \(2x+1\) lẻ \(\Rightarrow2x+1\inƯ\) lẻ của 36
\(\Rightarrow2x+1=\left\{\pm1;\pm3\right\}\)
\(TH1:2x+1=-1\Rightarrow x=-1\) và y không tồn tại ( loại )
\(TH2:2x+1=1\Rightarrow x=0\) và y không tồn tại ( loại )
\(TH4:2x+1=-3\Rightarrow x=-2\) và y không tồn tại ( loại )
\(TH4:2x+1=3\Rightarrow x=1\) và y không tồn tại ( loại )
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
Ta có :
\(36=6^2\)
\(\Rightarrow\left(2x-3\right)^2=6^2\)
\(\Rightarrow\left(2x-3\right)=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
:)
(2x -3) 2 = 36
(2x - 3)2 = 62
=> 2x - 3 = 6
=> 2x = 9
=> x = 9 : 2
=> x = 9/2 = 4,5