(2^3)*/6x+1=64
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15 tháng 7 2019
\(\left(\frac{1}{2}-\frac{1}{3}\right).6x+6x+2=67+64\)
\(\frac{\Rightarrow1}{6}.6x+6x+2=131\)
\(\Rightarrow x+6x=131-2\)
\(\Rightarrow7x=129\)
\(\Rightarrow x=\frac{129}{7}\)
1 tháng 6 2020
Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)
\(\Leftrightarrow\left(x+2\right)^3=\left(2x-1\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left[\left(x+2\right)-\left(2x-1\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(2x-1\right)+\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x+2-2x+1\right)\left(x^2+4x+4+2x^2+3x-2+4x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)
\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)
mà 7>0
nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)
Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)
AM
30 tháng 6 2015
a)6x2-12x=6x(x-2)
b)x2-12x+36=(x-6)2
c)8xy-16x2-y2=-(4x-y)2
d)125x3+1/64=(5x+1/4)(25x2-5/4x+1/16)
e)1/64x2+1/64=1/64(x2+1)
30 tháng 6 2015
a, 6x^2 - 12x = 6x ( x-2)
b, x^2 - 12x + 36 = x^2 - 2.x.6 + 36 = ( x -6)^2
c, 8xy - 16x^2 - y^2 = - [ 16x^2 - 8xy + y^2) = - (4x - y)^2
d, 125y^3 + 1/64 = ( 5y + 1/4)(25y^2 - 5/4y + 1/16)
e, 1/64 x^2 + 1/64 = 1/64( x^2+1)
29 tháng 11 2022
1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)
\(=8x^2+12x-8-8x^2+4x-16x+8\)
\(=0\)
3: \(=\left(6x+1-6x+1\right)^2=4\)
5: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x=9x\)
DH
8 tháng 7 2017
Bài 4:
a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)
\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)
\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)
\(=\left(x+4\right).\left(x+4\right)^2=\left(x+4\right)^3\)(1)
Thay \(x=6\) vào (1) ta được:
\(\left(6+4\right)^3=10^3=1000\)
Vậy...........
b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)
\(=x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2-4x+4\right)=\left(x-2\right).\left(x^2-2x-2x+4\right)\)
\(=\left(x-2\right).\left(x-2\right)^2=\left(x-2\right)^3\)(2)
Thay \(x=22\) vào (2) ta được:
\(\left(22-2\right)^3=20^3=8000\)
Vậy.............
Chúc bạn học tốt!!!
DH
8 tháng 7 2017
Bài 2:
a, \(\left(x+9\right)^3=27=3^3\)
\(\Rightarrow x+9=3\Rightarrow x=-6\)
Vậy.........
b, \(8-12x-x^3+6x^2=-64\)
\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)
\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)
\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)
\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)
Vậy............
Chúc bạn học tốt!!!
4 tháng 10 2021
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
KV
4 tháng 10 2023
[(6x² - 12) : 3] . 16 = 64
(6x² - 12) : 3 = 64 : 16
(6x² - 12) : 3 = 4
6x² - 12 = 4 . 3
6x² - 12 = 12
6x² = 12 + 12
6x² = 24
x² = 24 : 6
x² = 4
x = 2 hoặc x = -2
NM
4 tháng 10 2023
\(\left[\left(6x^2-12\right):3\right]\times16=64\\ \left(6x^2-12\right):3=4\\ 6x^2-12=12\\ 6x^2=24\\ x^2=4\\ x^2=2^2\\ x=2.\)
VN
1
12 tháng 2 2020
Đặt \(A=\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=18\)
\(B=\sqrt{x^2-6x+64}-\sqrt{x^2-6x+36}\)
\(\Rightarrow A.B=\left(x^2-6x+64\right)-\left(x^2-6x+36\right)=28\)
mà \(A=18\Rightarrow B=\frac{28}{18}=\frac{14}{9}\)
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(2 mũ 3 chia 4)nhân 2 mũ x+1bằng 64