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\(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow x-3=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
___________
\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3)
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27
<=> - 3x^2 - 18x - 22 = 0
<=> 3x^2 + 18x + 22 = 0
Nửa chu vi mảnh đất là:
120 : 2 = 60 (m)
Chiều dài hơn chiều rộng là:
5 + 5 = 10 (m)
Chiều rộng là:
( 60 - 10 ) : 2 = 25 (m)
Chiều dài là:
25 + 10 = 35 (m)
Diện tích là:
25 35 = 875 ( )
\(x^3-3x^2+3x-1=-8\)
\(\Leftrightarrow x-1=-2\)
hay x=-1
(3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6
18x+16=7
18x=7-16
x=-9/18=-2
vậy x =-2
1) \(\dfrac{3x}{4x-8}\)
\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)
2) \(\dfrac{2x}{x^2-9}\)
\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)
(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)
\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)
5) \(\dfrac{x-2}{x^2+3}\)
Do \(x^2+3>0\forall x\in R\)
Vậy biểu thức trên xác định với mọi x
6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)
\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
(3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
<=>(6x2+27x+4x+18)-(6x2+x+12x+2)=x+1-x+6
<=>6x2+31x+18-6x2-13x-2=7
<=>18x+16=7
<=>18x=-9
<=>x=-1/2
\(x^2-6x+9=\left(2x+1\right)^2\)
\(\Rightarrow\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Rightarrow\left(x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow\left(x-3-2x-1\right)\left(x-3+2x+1\right)=0\)
\(\Rightarrow-\left(x+4\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+4=0\\3x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=\frac{2}{3}\end{cases}}\)
x^2-6x+9=(2x+1)^2
<=>x2-6x+9=4x2+4x+1
<=>3x2+10x-8=0
<=>3x2-2x+12x-8=0
<=>x(3x-2)+4(3x-2)=0
<=>(x+4)(3x-2)=0
<=>x=-4 hoặc x=2/3