thu gon bieu thuc |2x+3|-x=1
giúp mk gấp nha
thanks
thu gon bieu thuc sau (2x-1)^3-(3x^2-1)(x-2)-(x+3)^3
Ta có: \(\left(2x-1\right)^3-\left(3x^2-1\right)\left(x-2\right)+\left(x+3\right)^3\)
\(=8x^3-12x^2+6x-1-\left(3x^3-6x^2-x+2\right)+x^3+9x^2+27x+27\)
\(=9x^3-3x^2+33x+26-3x^3+6x^2+x-2\)
\(=6x^3+3x^2+34x+24\)
x+2x+4x+...+125x+126x bang cach thu gon cac bieu thuc sau:
Đề không đúng
Tự dưng 1x+2x+4x+......+125x + 126x là sao
thu gon bieu thuc sau:
x+2x+3x+4x+...+125x+126
Cho bieu thuc :
C = ( 1 - 3x ) . x - 2x . ( x - 1 ) + 8
a ) Thu gon C
b ) Tinh gia tri bieu thuc voi x = -3
c ) Tim x de C = 8
d ) Tim ngiem cua C
dau . la dau nhan nha cac bn
nhanh nhanh nha mk dang can gap
thu gon bieu thuc
(x+1)^3-x(x^2+3)
giá trị rút gon cua bieu thuc
(2x-4)*(x+3)-2x(x+1)
(2x - 4)(x + 3) - 2x(x + 1)
= 2x2 + 2x - 12 - 2x(x + 1)
= 2x2 + 2x - 12 - 2x2 - 2x
= -12
Rut gon cacc bieu thuc sau : ( 4x - 3 ).( x - 5 ) - 2x ( 2x- 11)
\(\left(4x-3\right).\left(x-5\right)-2x\left(2x-11\right)\)
=\(4x^2-20x-3x+15-\left(4x^2-22x\right)\)
\(=4x^2-20x-3x+15-4x^2+22x\)
\(=-x+15\)
rut gon bieu thuc
(2x-1).(x-2)
Cho bieu thuc: ( x-1/ x+1 - x-1/x+1) : 2x / 3x - 3
a, Tim dieu kien xac dinh cua bieu thuc P
b, Rut gon bieu thuc P
c, Tim x thuoc z de P nhan gia tri nguyen.
Đề bài sai rồi bạn ! Mình sửa :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
b) \(P=\left(\frac{x-1}{x+1}-\frac{x+1}{x-1}\right):\frac{2x}{3x-3}\)
\(\Leftrightarrow P=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{x^2-2x+1-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-6}{x+1}\)
c) Để P nhận giá trị nguyên
\(\Leftrightarrow\frac{-6}{x+1}\inℤ\)
\(\Leftrightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)
Ta loại các giá trị ktm
\(\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)