(x^2-1)^2-18(x+1)(x-1)
a)Tìm x để B có gtnn:
B=x^2-2x+2011/x^2 (với x>0)
b) Giải pt:
1)1/x^2+2x-3+18/x^2+2x+2=18/x^2+2x+1
2) 8(x+1/x)^2+4(x^2+1/x^2)^2-4(x^2+1/x^2)(x+1/x)^2=(x+4)^2
\(1.\dfrac{x-1}{8}=\dfrac{2}{x-1}
\)
\(2.\dfrac{x+1}{2}=\dfrac{18}{x+1}\)
1. Tìm x:
a. 2x.(2x-1)^2 - 3x.(x+3).(x-3) - 4x.(x+1)^2 - x^3 = x^3 - 16.x^2 + 1
b. x.(x-2)^2 + 3x.(2x-1)^2 - 13.x^2 + 16.x^2 = 5x+2
2. Tính nhanh
a. 127^2 + 146 x 127 + 73^2
b. 9^8 x 2^8 - ( 18^4 - 1 ) x ( 18^4 + 1 )
help pls
1. Tìm giá trị của 1 x 5 x 18 + 2 x 10 x 36 + 3 x 15 x 54 / 1 x 3 x 9 + 2 x 6 x 18 + 3 x 9 x 27
[1-1/2] x [1-1/3] x [ 1-1/4] x .......x [1-1/18] x [ 1-1/19] x [ 1-1/2-]
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).........\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...........\frac{18}{19}.\frac{19}{20}=\frac{1.2.3.4.5........19}{2.3.4.5.6.7......20}=\frac{1}{20}\)
Tìm x 1) X/5=18/15 2)X/12=3/x 3) X - 2 1/2= 3 1/6 4) 2/5× x= 1/2
1: =>x/5=6/5
=>x=6
2: =>x^2=36
=>x=6 hoặc x=-6
3: =>x-5/2=3+1/6=19/6
=>x=19/6+5/2=19/6+15/6=34/6=17/3
4: =>x=1/2:2/5=1/2*5/2=5/4
a)|18-2x|=18
b)|1-x|=1-x
c)|x-2|=2-x
d)|x-2|=|3x+6|
a) \(\left|18-2x\right|=18\Rightarrow\orbr{\begin{cases}18-2x=18\\18-2x=-18\end{cases}\Rightarrow\orbr{\begin{cases}2x=0\\2x=36\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=18\end{cases}}}\)
b) \(\left|1-x\right|=1-x\Rightarrow\orbr{\begin{cases}1-x=1-x\\1-x=-\left(1-x\right)=-1+x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left\{\mp1;\mp2;...\right\}\\-x-x=-1-1\Rightarrow x=1\end{cases}}\)
Làm từng nãy đã , mỏi tay
a) ta có \(|\)18 - 2x\(|\)=18
=> 18-2x = 18 hoặc 18 -2x=-18
-2x = 0 -2x = -36
x = 0 x = 18
vậy x = 0 hoặc x=18
Giải phương trình:
a) \(\dfrac{x^2}{2}\) + \(\dfrac{18}{x^2}\) = 13(\(\dfrac{x}{2}\) - \(\dfrac{3}{x}\))
b) x(x - 1) + \(\dfrac{1}{x}\)(\(\dfrac{1}{x}\) - 1) = 0
a. ĐKXĐ:...
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}-3+3\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x}{2}-\dfrac{3}{x}\right)^2+6=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=t\Rightarrow2t^2-13t+6=0\Rightarrow\left[{}\begin{matrix}t=6\\t=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{3}{x}=6\\\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
\(\Leftrightarrow x\left(x-1\right)-\dfrac{x-1}{x^2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x=1\)
Tìm x: a)(2x+1)(1-2x)+(1-2x)^2=18
b) 2(x+1)^2-(x-3)(x+3)-(x-4)^2=0
c) (x-5)^2-x(x-4)=9
d) (x-5)^2+(x-4)(1-x)=0
a) (2x + 1)(1 - 2x) + (1 - 2x)2 = 18
= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18
= 2(1 - 2x) - 18 = 0
= 2 - 4x - 18 = 0
= -16 - 4x = 0
= -4x = 16
= x = \(\dfrac{16}{-4}=-4\)
b) 2(x + 1)2 -(x - 3)(x + 3) - (x - 4)2 = 0
= 2 (x2 + 2x + 1) - (x2 - 9) - (x2 - 8x + 16) = 0
= 2x2 + 4x + 2 - x2 + 9 - x2 + 8x - 16 = 0
= 12x - 5 = 0
= 12x = 5
= x = \(\dfrac{5}{12}\)
c) (x - 5)2 - x(x - 4) = 9
= x2 - 10x + 25 - x2 + 4x - 9 = 0
= -6x + 16 = 0
= -6x = -16
= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)
d) (x - 5)2 + (x - 4)(1 - x)
= x2 - 10x + 25 + 5x - x2 - 4 = 0
= -5x + 21 = 0
= -5x = -21
= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\)
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