Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
Suy ra: \(x^2-1+x-2x+1=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow12x=-6\)
hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
p) \(\left(9-x\right)\left(x^2+2x-3\right)\)
\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)
\(=9x^2+18x-27-x^3-2x^2+3x\)
\(=-x^3+7x^2+21x-27\)
n) \(\left(-x+3\right)\left(x^2+x+1\right)\)
\(=-x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)
\(=-x^3-x^2-x+3x^2+3x+3\)
\(=-x^2+2x^2+2x+3\)
o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)
\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)
\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)
\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)
q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)
\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=6x^3-12x^2-18x+x^2-2x-3\)
\(=6x^3-11x^2-20x-3\)
r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)
\(=2x\left(-x^2-3x+1\right)+\left(-x^2-3x+1\right)\)
\(=-2x^3-6x^2+2x-x^2-3x+1\)
\(=-2x^3-7x^2-x+1\)
u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)
\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)
\(=-2x^3+2x^2+12x+3x^2-3x-18\)
\(=-2x^3+5x^2+9x-18\)
s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)
\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)
\(=-4x^3-12x^2+8x+5x^2+15x-10\)
\(=-4x^3-7x^2+23x-10\)
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)
\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)
\(=-x^2-3+2x^4+6x+18-12x^3\)
\(=2x^4-12x^3-x^2+6x+15\)
p: (-x+9)(x^2+2x-3)
=-x^3-2x^2+3x+9x^2+18x-27
=-x^3+7x^2+21x-27
n: (-x+3)(x^2+x+1)
=-x^3-x^2-x+3x^2+3x+3
=-x^3+2x^2+2x+3
o: (-6x+1/2)(x^2-4x+2)
=-6x^3+24x^2-12x+1/2x^2-2x+1
=-64x^3+49/2x^2-14x+1
q: (6x+1)(x^2-2x-3)
=6x^3-12x^2-18x+x^2-2x-3
=6x^3-11x^2-20x-3
r: (2x+1)(-x^2-3x+1)
=-2x^3-6x^2+2x-x^2-3x+1
=-2x^3-7x^2-x+1
u: =-2x^3+2x^2+12x+3x^2-3x-18
=-2x^3+5x^2+9x-18
s: =-4x^3-12x^2+8x+5x^2+15x-10
=-4x^3-7x^2+23x-10
c: \(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=4x^2+12x+9+4x^2-12x+9-\left(4x^2-9\right)\)
\(=8x^2+18-4x^2+9=4x^2+27\)
d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)
\(=x^3-1-8x^3-27=-7x^3-28\)
e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)
\(=x^3+3x^2+3x+1-6x^2-\left(x^3-3x^2+3x-1\right)\)
\(=x^3-3x^2+3x+1-x^3+3x^2-3x+1\)
=2
1.A =( x-3)( x+3) + 15 - x2
A=X2-3X+3X+15-X3
A=15-X
2.B=(X -1) (X2+X+1) - X (X2+2) + 2X
B=X3+ X2+ X - X2 - X - 1 - X3 - 2X + 2X
B= -1
3.C=(2X - 1 ) (4X2 + 2X + 1) - X ( 8 X 2 + 1 ) + X
C=8X3 - 4X2 +4X2 - 2X +2 X - 1 - 8X22 - X + X
C=8X3 - 1 - 8X22
MK CHỈ LM ĐC TỚI ĐÓ THUI SAI CHỖ NÀO ĐỪNG TRÁCH VÌ MK YẾU PHẦN NÀY
Lời giải:
a.
a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$
$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$
$\Leftrightarrow 3x+1=5x-3$
$\Leftrightarrow 4=2x$
$\Leftrightarrow x=2$
b.
$(2x-1)(x+3)-(x-2)(x+3)=3x+1$
$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$
$\Leftrightarrow x^2+5x+1=3x+1$
$\Leftrightarrow x^2+2x=0$
$\Leftrightarrow x(x+2)=0$
$\Leftrightarrow x=0$ hoặc $x=-2$
c.
$x^2(x-1)-x(x-1)(x+1)=0$
$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$
$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$
$\Leftrightarrow (x-1)(-x)=0$
$\Leftrightarrow x-1=0$ hoặc $-x=0$
$\Leftrightarrow x=1$ hoặc $x=0$
d.
$4x(x-5)-(2x-3)(2x+3)=9$
$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$
$\Leftrightarrow -20x=0$
$\Leftrightarrow x=0$
a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow2x=4\)
hay x=2
b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)
\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)
\(\Leftrightarrow4x^2-20x-4x^2+9=9\)
hay x=0
\(\dfrac{x+1}{x-3}\) + \(\dfrac{x-1}{x+3}\) - \(\dfrac{2x-2x^2}{9-x^2}\)
= \(\dfrac{x+1}{x-3}\)+ \(\dfrac{x-1}{x+3}\) + \(\dfrac{2x-2x^2}{x^2-9}\)
= \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\) + \(\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\) + \(\dfrac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+3x+x+3+\left(x^2-3x-x+3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+4x+3+x^2-3x-x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)