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a) (x + 5)2 - (x - 3)2 = 2x - 7
(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7
8(2x + 2)= 2x - 7
16x + 16 = 2x - 7
16x - 2x = - 7 - 16
14x = - 23
x = - 23/14
b) (2x - 3)(4x2 + 6x + 9) = 98
(2x)3 - 33 = 98
8x3 - 27 = 98
8x3 = 125
x3 = 125/8
x3 = (5/2)3
x = 5/2
a) A =(3x+7)(2x+3)-(3x-5)(2x+11)
=6x2+9x+14x+21-(6x2+33x-10x-55
=6x2+9x+14x+21-6x2-33x+10x+55
=76
vậy A ko phạu thuộc vào giá trị của x
b)B =(x2-2)(x2+x-1)-x(x3+x2-3x-2)
=x4+x3-x2-2x2-2x+2-x4-x3+3x2+2x
=2
vậy giá trị của B ko phụ thuộc vào x
a, \(\dfrac{x+1}{3}+\dfrac{2x-1}{3}=\dfrac{x+1+2x-1}{3}=\dfrac{3x}{3}=x\)
b, \(\dfrac{5x-2y}{x^2-y^2}+\dfrac{y-4x}{x^2-y^2}=\dfrac{5x-2y+y-4x}{\left(x+y\right)\left(x-y\right)}=\dfrac{x-y}{\left(x+y\right)\left(x-y\right)}=\dfrac{1}{x+y}\)
c, \(\dfrac{x-1}{12x}+\dfrac{2x+7}{12x}+\dfrac{6-3x}{12x}=\dfrac{x-1+2x+7+6-3x}{12x}=\dfrac{12}{12x}=\dfrac{1}{x}\)
a,\(\dfrac{x+1_{ }+2x-1}{3}\)=\(\dfrac{2x}{3}\)
b,\(\dfrac{5x-2y+y-4x}{\left(x+y\right)\left(x-y\right)}\)=\(\dfrac{x-y}{\left(x+y\left(x-y\right)\right)}\)
c,\(\dfrac{x-1+2x+7+6-3x}{12x}\)=\(\dfrac{12}{12x}\)=\(\dfrac{1}{x}\)
Mình nghĩ đề câu a là: \(x+\sqrt{5}+\sqrt{x}-1=-6\)
Đặt \(\sqrt{x}=t\Rightarrow t^2=x\)
\(Ta\)\(được\): \(t^2+\sqrt{5}+t-1=-6\)
\(\Leftrightarrow t^2-5+t+\sqrt{5}=0\)
\(\Leftrightarrow\left(t-\sqrt{5}\right).\left(t+\sqrt{5}\right)+\left(t+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left(t+\sqrt{5}\right).\left(t-\sqrt{5}+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}t=-\sqrt{5}\\t=\sqrt{5}-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=5\\x=6-2\sqrt{5}\end{cases}}\)
a: Ta có: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{x^2-4y^2}\)
\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(\)\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)
b: \(\frac{4x+7}{2x+2}-\frac{3x+6}{2x+2}\)
\(=\frac{4x+7-3x-6}{2x+2}\)
\(=\frac{x+1}{2\left(x+1\right)}=\frac12\)
c: \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)
\(=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\cdot\left(x+3\right)}\)
\(=\frac{x\left(x+9\right)-3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)
d: \(\frac{1}{x^2+3x+2}-\frac{1}{x^2-4}\)
\(=\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-2-\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{-3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)