(2x2-3)3=27/8
Giúp mk với đi
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xin lỗi mk viết thiếu => 3A-A= 19683-1
=> 2A=19683-1
=> A= (19683-1) :2
= 9841
A = 1 + 3 + 9 + 27 + ... + 2187
A x 3 = 1 x 3 + 3 x 3 + ... + 2187 x 3
A x 3 = 3 + 9 + ... + 6561
A = 6561 - 3 = 6558
Có \(\left(x-\dfrac{1}{3}\right)^2\ge0\)
Mà \(\left(x-\dfrac{1}{3}\right)^2=-\dfrac{8}{27}< 0\) \(\Rightarrow\) Vô lý
\(2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
Đặt : \(A=2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+0\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-40\right]\right\}\)
\(A=2017-\left\{25.4-11.9\right\}\)
\(A=2017-\left\{25.4-99\right\}\)
\(A=2017-\left\{100-99\right\}\)
\(A=2017-1=2016\)
Vậy A = 2016
a)\(12^3.3^3\)
\(=\left(12.3\right)^3\)
\(=36^3\)
\(=46656\)
b)\(2.8^4\)
\(=2.\left(2^3\right)^3\)
\(=2.2^9\)
\(=2^{10}\)
\(=1024\)
a: =>2x^2+9x-6x-27=0
=>x(2x+9)-3(2x+9)=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
b: =>-10x^2+6x-5x+3=0
=>-2x(5x-3)-(5x-3)=0
=>(5x-3)(-2x-1)=0
=>x=-1/2 hoặc x=5/3
c: =>-x^3+2x^2-x^2+4=0
=>-x^2(x-2)-(x-2)(x+2)=0
=>(x-2)(-x^2-x-2)=0
=>x-2=0
=>x=2
d: =>(x^3+8)-4x(x+2)=0
=>(x+2)(x^2-2x+4)-4x(x+2)=0
=>(x+2)(x^2-6x+4)=0
=>x=-2 hoặc \(x=3\pm\sqrt{5}\)
1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
=>x=0
2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5=7\)
=>6x-12=7
=>6x=19
hay x=19/6
(2x2 - 3)3=\(\frac{27}{8}\)
<=> (2x2 - 3)3=\(\left(\frac{3}{2}\right)^3\)
<=> 2x2 - 3 = \(\frac{3}{2}\)
<=> 2x2 =\(\frac{3}{2}+3\)
<=>2x2=\(\frac{9}{2}\)
<=> x2 = \(\frac{9}{2}:2\)
<=> x2=\(\frac{9}{4}\)
\(\Rightarrow x\in\left\{-\frac{3}{2};\frac{3}{2}\right\}\)
vậy...
\(\left(2x^2-3\right)^3=\frac{27}{8}\)
\(\Leftrightarrow2x^2-3=\frac{3}{2}\)
\(\Leftrightarrow2x^2=\frac{9}{2}\)
\(\Leftrightarrow x^2=\frac{9}{4}\)
\(\Leftrightarrow x=\pm\frac{3}{2}\)