\(|\)x-1,5\(|\)= 2
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\(\left(x-1,5\right)^6+2.\left(1,5-x\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\1,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow x=1,5\)
Vậy : \(x=1,5\)
a \(\dfrac{3}{8}\times\dfrac{5}{2}+\dfrac{5}{8}\times\dfrac{5}{2}=\dfrac{5}{2}\left(\dfrac{3}{5}+\dfrac{5}{8}\right)=\dfrac{5}{2}\times\dfrac{8}{8}=\dfrac{5}{2}\times1=\dfrac{5}{2}\)
b \(33.12\times0=0\)
\(\left(x-1,5\right)^6+2.\left(1,5-x\right)^2=0\\ \Leftrightarrow\left(x-1,5\right)^6+2.\left(x-1,5\right)^2=0\\ \Leftrightarrow\left(x-1,5\right)^2.\left[\left(x-1,5\right)^4+2\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1,5\right)^2=0\\\left(x-1,5\right)^4+2\ge0\forall x\in R\end{matrix}\right.\\ \Leftrightarrow x=1,5\)
Vậy x=1,5
Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
Làm:
\(1,5\cdot x-\frac{7}{3}x=1,5-\frac{2}{3}\)
\(\left(1,5-\frac{7}{3}\right)x=\frac{5}{6}\)
\(\Leftrightarrow\) \(-\frac{5}{6}x=\frac{5}{6}\)
\(\Leftrightarrow\) \(x=-1\)
~ học tốt ~
1,5 . x - 7/3 . x = 1,5 - 2/3
=> 1,5 . x - 7/3 . x = 5/6
=>1,5 . x = 5/6 - 7/3
=> 1,5 . x = - 3/2
=> x = - 3/2 : 1,5
=> x = -1
Vay x = - 1
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}
27,5 × 2 + 22,5 × 2 =(27,5 + 22,5) × 2 = 50 × 2 = 100
167,77 × 5 + 132,23 × 5 = (167,77 + 132,23) × 5 = 300 × 5 = 1500
1,5 × 10 + 1,5 × 10 = (1,5 + 1,5) × 10 = 3 × 10 = 30
10,2 × 2 + 9,8 × 2 = (10,2 + 9,8) × 2 = 20 × 2 = 40
a. 168x168-168x68
= 168x(168-68)
= 168x100
= 16800
b. 201,5x17-2015x16-201,5
= 201,5x17-201,5x160-201,5
= 201,5x(17-16-1)
= 201,5x0
= 0
c. 1,5 + 1,5x2+1,5x7-10
= 1,5x1+1,5x2+1,5x7-10
= 1,5x(1+2+7)-10
= 1,5x10-10
= 10x(1,5-1)
= 10x0,5
= 5
d. 1,25x2,5x64
= 1,25x64x2,5
=80x2,5
= 200
a)168*168-168*68=168*(168-68)=168*100=16 800
b)201,5*17-2015*16-201,5=201,5*17-201,5*160-201,5=201,5*(17-160-1)=201,5*(-144)=-29016
c)1,5+1,5*2+1,5*7-10=1,5*(1+2+7)-10=1,5*10-10=(1,5-1)*10=0,5*10=5
d)1,25*2,5*64=1,25*2,5*8*8=(1,25*8)*(2,5*8)=10*20=200
bn ơi bài này làm 2 trường hợp
\(\left|x-1,5\right|=2\\ x-1,5=2\\ x=0,5\)