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1.
a) \(x\left(1-x\right)+\left(x+1\right)\left(x-2\right)\)
\(=x-x^2+x^2-2x+x-2\)
\(-2\)
b) \(\left(x+3\right)^2-x^2=45\)
\(x^2+6x+9-x^2=45\)
\(6x+9=45\)
\(6x=36\)
\(x=6\)
Bài làm
a) x( 1 - x ) + ( x + 1 )( x - 2 )
= x - x2 + x2 - 2x + x - 2
= ( x - 2x + x ) + ( x2 - 2x ) - 2
= -2
b) ( x + 3 )2 - x2 = 45
=> ( x + 3 - x )( x + 3 + x ) = 45
=> 3. ( 2x + 3 ) = 45
=> 6x + 9 = 45
=> 6x = 45-9
=> 6x = 36
=> x = 6
Vậy x = 6
# Học tốt #
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
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: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
\(\text{ĐKXĐ:}\hept{\begin{cases}x\ne1\\x\ne2\\x\ne3\end{cases}}\)
\(\frac{x+2}{\left(x-2\right)\left(x-3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}:\frac{2x-2-x}{x-1}\)
\(=\frac{x+2+x^2-9+x^2-4}{\left(x-2\right)\left(x-3\right)}.\frac{x-1}{x-2}=\frac{2x^2+x-11}{\left(x-2\right)\left(x-3\right)}\cdot\frac{x-1}{x-2}=\frac{\left(x-1\right)\left(2x^2+x-11\right)}{\left(x-2\right)^2\cdot\left(x-3\right)}\)
\(C=\left(1-\frac{x}{x+1}\right)\div\left(\frac{x+3}{x-2}+\frac{2+x}{3-x}+\frac{x+2}{x^2-5x+6}\right)\)
ĐKXĐ : x ≠ -1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 11/5
\(=\left(\frac{x+1}{x+1}-\frac{x}{x+1}\right)\div\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\left(\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-4}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\frac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{1}{x+1}\times\frac{x-2}{1}\)
\(=\frac{x-2}{x+1}\)
Đặt \(A=\frac{x^2+x-6}{x^3-4x^2-18x+9}\)
\(A=\frac{x^2+3x-2x-6}{x^3+3x^2-7x^2-21x+3x+9}\)
\(A=\frac{x\left(x+3\right)-2\left(x+3\right)}{x^2\left(x+3\right)-7x\left(x+3\right)+3\left(x+3\right)}\)
\(A=\frac{\left(x-2\right)\left(x+3\right)}{\left(x^2-7x+3\right)\left(x+3\right)}\)
\(A=\frac{x-2}{x^2-7x+3}\)
P = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
P = \(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Với \(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)
=> P = \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-3}=\frac{\sqrt{5}-1+1}{\sqrt{5}-1-3}=\frac{\sqrt{5}}{\sqrt{5}-4}=\frac{\sqrt{5}\left(\sqrt{5}+4\right)}{\left(\sqrt{5}-4\right)\left(\sqrt{5}+4\right)}=\frac{5+4\sqrt{5}}{-11}\)