Tìm x biết a, A= (x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2 = 15
toán 8 mn ạ, mong có thể trình bày rõ ràng ko tắt giúp em ;-;
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(4\left(x-2\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
1) 1/3 x 1/2 x 3/7 = 3/42 = 1/14
2) 5/4 x 1/3 +1/7 = 5/12 + 1/7 = 35/84 + 12/84 = 47/84
3) 8 x ( 8/9 - 2/3 ) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 240/240 = 1
5) ( 2/5 + 3/4 ) + 8 = 23/20 + 8 = 23//20 + 160/20 = 183/20
6) 10 x ( 1/2 - 1/5 ) = 10 x 3/10 = 10/1 x 3/10 = 30/10 = 3
Ta có: \(C=\left(\dfrac{x}{2}-y\right)^3-6\left(y-\dfrac{x}{2}\right)^2+12\left(y-\dfrac{x}{2}\right)-8\)
\(=\left(\dfrac{x}{2}-y\right)^3-3\cdot\left(\dfrac{x}{2}-y\right)^2\cdot2-3\cdot\left(\dfrac{x}{2}-y\right)\cdot2^2-2^3\)
\(=\left(\dfrac{x}{2}-y\right)^3-8-6\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}-y-2\right)\)
\(=\left(\dfrac{x}{2}-y-2\right)\left[\left(\dfrac{x}{2}-y\right)^2+2\left(\dfrac{x}{2}-y\right)+2^2\right]-6\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}-y-2\right)\)
\(=\left(\dfrac{x}{2}-y-2\right)\left[\left(\dfrac{x}{2}-y\right)^2-4\left(\dfrac{x}{2}-y\right)+4\right]\)
\(=\left(\dfrac{x}{2}-y-2\right)^3\)
\(\Rightarrow\left(x+1\right)\times\left(x-1+1\right):2=1225\\ \Rightarrow\left(x+1\right)\times x=1225\times2=2450=49\times50\\ \Rightarrow x=49\)
\(2x-1-x^2\\ =x+x-1-x^2\\ =\left(x-x^2\right)+\left(x-1\right)\\ =-x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(1-x\right)\)
x.(x+3).(3x+6)=10
x.x+x.3.(3x+6)=10
x mũ 2+3x.3x+6=10
x mũ 2+3x mũ 2+6=10
x mũ 2+3x mũ 2+6=10
(3x+x) mũ 2=10-6
(3x+x) mũ 2=4
4x mũ 2=4
=> x mũ 2=1
=>x=1
nhớ tick đúng nhé
\(\left|x+6\right|-9=2x\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+6-9=2x\\x-6+9=2x\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2x=-6+9\\x-2x=6-9\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}-x=3\\-x=-3\end{array}\right.\)
Vậy \(x=-3\)
Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)