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bài 1 .
a. 3 x(5x2 – 2x -1) = 15x3 – 6x2 – 3x
b. (x2+2xy -3)(-xy) = – x3y – 2x2y2 + 3xy
c. 1/2 x2y ( 2x3 – 2/5 xy2 -1 )= x5y – 1/5 x3y3 – 1/2 x2y
bài 2 .
a) 2x^3-3x-5x^3-x^2+x^2=-3x-3x^3
b) 3x^2-6x-5x+5x^2-8x^2+24=-11x+24
c) 3x^3-3/2x^2-x^3-x/2+x/2+2=2x^3-3/2x^2+2
bài 3 .
?????????? bài 3 thì tui ko biết
Bài 3 :
\(P=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(=5x^3-15x+7x^2-5x^3-7x^2=-15x\)
Thay x = -5 vào biểu thức trên ta được
\(-15.\left(-5\right)=75\)
Vậy x = -5 thì P = 75
|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}
a) \(x^2-1+\left(3-x\right)x=5\Leftrightarrow x^2-1+3x-x^2=5\)
\(\Leftrightarrow3x-1=5\Leftrightarrow3x=6\Leftrightarrow x=\dfrac{6}{3}=2\) vậy \(x=2\)
b) \(7\left(2x-5\right)-5\left(7x-2\right)+2\left(5x-7\right)=6\)
\(\Leftrightarrow14x-35-35x+10+10x-14=6\)
\(\Leftrightarrow-11x-39=6\) \(\Leftrightarrow-11x=45\Leftrightarrow x=-\dfrac{45}{11}\) vậy \(x=-\dfrac{45}{11}\)
c) \(\dfrac{2x-1}{3}+\dfrac{x-2}{2}=1\Leftrightarrow\dfrac{4x-2}{6}+\dfrac{3x-6}{6}=1\)
\(\Leftrightarrow\dfrac{4x-2+3x-6}{6}=1\Leftrightarrow4x-2+3x-6=6\)
\(\Leftrightarrow7x-8=6\Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\) vậy \(x=2\)
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
Có: \(f\left(x\right)=5x^3-7x^2+x+7\)
\(g\left(x\right)=7x^3+2x+5\)
\(h\left(x\right)=2x^3+4x+1\)
\(\Rightarrow k\left(x\right)=f\left(x\right)-g\left(x\right)+h\left(x\right)\\ =5x^3-7x^2+x+7-7x^3-2x-5+2x^3+4x+1\\ =\left(5x^3-7x^3+2x^3\right)+\left(7-5+1\right)+\left(4x-2x+x\right)-7x^2\\ =0+3+3x-7x^2\\ =-7x^2+3x+3\)
Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết 
1, \(\left(x-4\right)^2-\left(2x+1\right)^2=\left(x-4-2x-1\right)\left(x-4+2x+1\right)=-3\left(x+5\right)\left(x-1\right).\)
\(\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)(mấy cái này áp dụng hàng đẳng thức lớp 8 mới hok)
2,\(x^3+x^2-4x-4=\left(x-2\right)\left(x^2+3x+2\right)=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(\orbr{\begin{cases}x=\mp2\\\end{cases}}x=-1\)
tương tụ lm tiếp nhe buồn ngủ quá rồi !
