Rút gọn phân thức
a) \(\frac{x^3+3x^2-4}{x^3-3x+2}\)
b) \(\frac{x^3+^2-4x-4}{x^3+8x^2+17x+10}\)
rút gọn
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
\(=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+x^2+7x^2+7x+10x+10}\)
\(=\frac{\left(x^2-4\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{x^2-4}{x^2+7x+10}\)
\(=\frac{x^2-4}{x^2+5x+2x+10}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+5\right)+2\left(x+5\right)}\)
\(=\frac{x-2}{x+5}\)
Rút gọn phân thức:
\(a,\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(b,\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
Còn bài b bạn tự làm nhé
Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)
Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
bài 1 : Rút gọn biểu thức
a) 3x.(x-2)-5x(1-x)-8(x^2-3)
b) (7x-3) (2x+1) - (5x-2) (x+4)-9x^2 + 17x
\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=0-11x+24\)
\(=-11x+24\)
\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)
\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)
\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)
\(=0+0+5\)
\(=5\)
Rút gọn phân thức
a) x^3 -x / 3x + 3
b) x^2 + 3xy / x^2 - 9y^2
c) x^2 + 4x + 4 / 3x + 6
\(a,\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\\ b,\dfrac{x^2+3xy}{x^2-9y^2}=\dfrac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\dfrac{x}{x-3y}\\ c,\dfrac{x^2+4x+4}{3x+6}=\dfrac{\left(x+2\right)^2}{3\left(x+2\right)}=\dfrac{x+2}{3}\)
Tìm đkxđ rồi rút gọn phân thức sau:
A= \(\frac{x^3-3x^2+x+3}{x^3-x^2}\)
B=\(\frac{4x^3-8x^2+3x-6}{12x^3+4x^2+9x+3}\)
C=\(\frac{x^4-1}{x^3+2x^2-x-2}\)
D=\(\frac{a^3+b^3}{a^3+\left(a-b\right)^3}\)
a, Điều kiện xác định: x<>0
b, Điều kiện xác định: x <> -1/3
c, Điều kiện xác định: x<>2
d, Điều kiện xác định: a<>0 và b<>0; b<>2a
A : không rút gọn được
\(B=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{3x\left(4x^2+3\right)+4x^2+3}=\frac{\left(4x^2+3\right)\left(x-2\right)}{\left(4x^2+3\right)\left(3x+1\right)}=\frac{x-2}{3x+1}\)
\(C=\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
\(D=\frac{a^3+b^3}{a^3+\left(a-b\right)^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+a-b\right)\left(a^2-a^2+ab+a^2-2ab+b^2\right)}\)\(=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(2a-b\right)\left(a^2-ab+b^2\right)}=\frac{a+b}{2a-b}\)
cho \(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-8x+16}{x-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)
rút gọn A
Cho các biểu thức
A = \(\dfrac{1}{x+2}-\dfrac{2x}{4-x^2}+\dfrac{3}{x-2}\) và B = \(\dfrac{x+2}{3x+2}\)với x ≠ 2; x ≠ -2; x ≠ -\(\dfrac{2}{3}\)
a. Tính giá trị của A biết \(3x^2+8x+4=0\)
b. Rút gọn B
\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
Rút gọn các phân thức
a)\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
b) \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)
a) \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
= \(\dfrac{\left(x^3+x^2\right)-\left(4x+4\right)}{\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)}\)
=\(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
= \(\dfrac{\left(x^2-4\right)\left(x+1\right)}{\left(x^2+7x+10\right)\left(x+1\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)\left(x+1\right)}{\left[\left(x^2+2x\right)+\left(5x+10\right)\right]\left(x+1\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+5\right)\left(x+2\right)}\)
= \(\dfrac{x-2}{x+5}\)
b) \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)
= \(\dfrac{x^3-x^2+4x^2+4x-4x-4}{x^3-x-2x+2}\)
= \(\dfrac{\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)}{\left(x^3-x\right)-\left(2x-2\right)}\)
= \(\dfrac{x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)}{x^2\left(x-1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x^2+4x+4\right)\left(x-1\right)}{\left(x^2-2\right)\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2\left(x-1\right)}{\left(x^2-2\right)\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2}{x^2-2}\)
Nhớ tik nha...
chữa lại câu b nhé:
b) \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)
= \(\dfrac{x^3-x^2+4x^2+4x-4x-4}{x^3-x-2x+2}\)
=\(\dfrac{\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)}{\left(x^3-x\right)-\left(2x-2\right)}\)
= \(\dfrac{x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)}{x\left(x^2-1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x^2+4x+4\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2\left(x-1\right)}{\left(x-1\right)\left[x\left(x+1\right)-2\right]}\)
= \(\dfrac{\left(x+2\right)^2}{x^2+x-2}\)
Đúng chưa bn...
bài 1 rút gọn biểu thức
a) (2x-5)^2-4x(x+3)
b) (x-2)^3 -6(x+4)(x-4)-(x-2)(x^2+2x+4)
c)(x-1)^2-2(x-1)(x+2)+(x+2)^2+5(2x-3)
bài 2 rút gọn biểu thức
a)(2-3x)^2-5x(x-4)+4(x-1)
b)(3-x)(x^2+3x+9)+(x-3)^3
c)(x-4)^2(x+4)-(x-4)(x+4)^2+3(x^2-16)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)