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a/ \(x^2-6x+5=0\)
\(x^2-5x-x+5=0\)
\(x\left(x-5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}x-5=0\rightarrow x=5\\x-1=0\rightarrow x=1\end{cases}}\)
b/\(2x^2+7x+9=0\)
?!
c/ \(4x^2-7x+3=0\)
\(4x^2-4x-3x+3=0\)
\(4x\left(x-1\right)-3\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x-3\right)=0\)
\(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\4x-3=0\Rightarrow x=\frac{3}{4}\end{cases}}\)
d/ \(2\left(x+5\right)=2x+10\)
-,- mik ko rõ đề ạ, sai thì ibox ạ.Cảm ơn
x2 - 5x - 36 = 0
=> x2 - 9x + 4x - 36 = 0
=> x(x - 9) + 4(x - 7) = 0
=> (x + 4)(x - 7) = 0
=> \(\orbr{\begin{cases}x+4=0\\x-7=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)
6x2 - (2x + 5)(3x - 2) = -12
=> 6x2 - 6x2 + 4x - 15x + 10 = -12
=> -11x = -22
=> x = 2
x2 - 25 = 6x - 9
=> x2 - 25 - 6x + 9 = 0
=> x2 - 6x - 16 = 0
=> x2 - 8x + 2x - 16 = 0
=> x(x - 8) + 2(x - 8) = 0
=> (x + 2)(x - 8) = 0
=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
A, => (x-2)(x+2)(x-3)(x-3)=0
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+2=0\\x-3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\\x=3\\x=-3\end{matrix}\right.\)
Vậy tập no của pt ..
B, \(\Leftrightarrow x^2+2x+5x+10=0\)
\(\Leftrightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-2\\x=-5\end{matrix}\right.\)
Vậy ...............
C, \(\Leftrightarrow3x^2-6x-6x+12=0\)
\(\Leftrightarrow3x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\3x-6=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=2\end{matrix}\right.\)
Vậy tập no của pt là S{2}
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)
\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)
\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)
\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)
`@` `\text {Ans}`
`\downarrow`
`1.`
\(\left(-4xy\right)\cdot\left(2xy^2-3x^2y\right)\)
`=`\(\left(-4xy\right)\left(2xy^2\right)+\left(-4xy\right)\left(-3x^2y\right)\)
`=`\(-8\left(x\cdot x\right)\left(y\cdot y^2\right)+12\left(x\cdot x^2\right)\left(y\cdot y\right)\)
`=`\(-8x^2y^3+12x^3y^2\)
`2.`
\(\left(-5x\right)\left(3x^3+7x^2-x\right)\)
`=`\(\left(-5x\right)\left(3x^3\right)+\left(-5x\right)\left(7x^2\right)+\left(-5x\right)\left(-x\right)\)
`=`\(-15x^4-35x^3+5x^2\)
`3.`
\(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)
`=`\(3x\left(4x+5\right)-2\left(4x+5\right)-12x^2+6x\)
`=`\(12x^2+15x-8x-10-12x^2+6x\)
`=`\(\left(12x^2-12x^2\right)+\left(15x-8x+6x\right)-10\)
`=`\(13x-10\)
`4.`
\(2x^2\left(x^2-7x+9\right)\)
`=`\(2x^2\cdot x^2+2x^2\cdot\left(-7x\right)+2x^2\cdot9\)
`=`\(2x^4-14x^3+18x^2\)
`5.`
\(\left(3x-5\right)\left(x^2-5x+7\right)\)
`=`\(3x\left(x^2-5x+7\right)-5\left(x^2-5x+7\right)\)
`=`\(3x^3-15x^2+21x-5x^2+25x-35\)
`=`\(3x^3-20x^2+46x-35\)
8)
\(\frac{x + 9}{x^{2} - 9} - \frac{3}{x^{2} + 3 x}\)
\(= \frac{x + 9}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} - \frac{3}{x \left(\right. x + 3 \left.\right)}\)
\(= \frac{x \left(\right. x + 9 \left.\right) - 3 \left(\right. x - 3 \left.\right)}{x \left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\)
\(= \frac{x^{2} + 9 x - 3 x + 9}{x \left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\)
\(= \frac{x^{2} + 6 x + 9}{x \left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\)
\(= \frac{\left(\right. x + 3 \left.\right)^{2}}{x \left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\)
\(= \frac{x + 3}{x \left(\right. x - 3 \left.\right)}\)
9)
\(\frac{x - 12}{6 x - 36} + \frac{6}{x^{2} - 6 x}\)
\(= \frac{x - 12}{6 \left(\right. x - 6 \left.\right)} + \frac{6}{x \left(\right. x - 6 \left.\right)}\)
\(= \frac{x \left(\right. x - 12 \left.\right) + 36}{6 x \left(\right. x - 6 \left.\right)}\)
\(= \frac{x^{2} - 12 x + 36}{6 x \left(\right. x - 6 \left.\right)}\)
\(= \frac{\left(\right. x - 6 \left.\right)^{2}}{6 x \left(\right. x - 6 \left.\right)}\)
\(= \frac{x - 6}{6 x}\)
10)
\(\frac{3 x + 5}{x^{2} - 5 x} + \frac{25 - x}{25 - 5 x}\)
\(= \frac{3 x + 5}{x \left(\right. x - 5 \left.\right)} - \frac{25 - x}{5 \left(\right. x - 5 \left.\right)}\)
\(= \frac{5 \left(\right. 3 x + 5 \left.\right) - x \left(\right. 25 - x \left.\right)}{5 x \left(\right. x - 5 \left.\right)}\)
\(= \frac{15 x + 25 - 25 x + x^{2}}{5 x \left(\right. x - 5 \left.\right)}\)
\(= \frac{x^{2} - 10 x + 25}{5 x \left(\right. x - 5 \left.\right)}\)
\(= \frac{\left(\right. x - 5 \left.\right)^{2}}{5 x \left(\right. x - 5 \left.\right)}\)
\(= \frac{x - 5}{5 x}\)
8) \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)
\(=\frac{x\left(x+9x\right)}{x\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x+3}{x\left(x-3\right)}\)
9) \(\frac{x-12}{6x-36}+\frac{6}{x^2-6x}\)
\(=\frac{x\left(x-12\right)}{6x\left(x-6\right)}+\frac{36}{6x\left(x-6\right)}\)
\(=\frac{x^2-12x+36}{6x\left(x-6\right)}\)
\(=\frac{\left(x-6\right)^2}{6x\left(x-6\right)}\)
\(=\frac{x-6}{6x}\)
10) \(\frac{3x+5}{x^2-5x}+\frac{25-x}{25-5x}\)
\(=\frac{3x+5}{x\left(x-5\right)}-\frac{25-x}{5\left(x-5\right)}\)
\(=\frac{5\left(3x+5\right)}{5x\left(x-5\right)}-\frac{x\left(25-x\right)}{5x\left(x-5\right)}\)
\(=\frac{15x+25-25x+x^2}{5x\left(x-5\right)}\)
\(=\frac{x^2-10x+25}{5x\left(x-5\right)}\)
\(=\frac{\left(x-5\right)^2}{5x\left(x-5\right)}\)
\(=\frac{x-5}{5x}\)
Bài 8) $\frac{x+9}{x^2-9} - \frac{3}{x^2+3x}$
Nhận thấy tử số $x^2 + 6x + 9 = (x + 3)^2$:$$= \frac{(x + 3)^2}{x(x - 3)(x + 3)} = \frac{x + 3}{x(x - 3)}$$
Bài 9) $\frac{x-12}{6x-36} + \frac{6}{x^2-6x}$
Nhận thấy tử số $x^2 - 12x + 36 = (x - 6)^2$:$$= \frac{(x - 6)^2}{6x(x - 6)} = \frac{x - 6}{6x}$$
Bài 10) $\frac{3x+5}{x^2-5x} + \frac{25-x}{25-5x}$
Nhận thấy tử số $x^2 - 10x + 25 = (x - 5)^2$:$$= \frac{(x - 5)^2}{5x(x - 5)} = \frac{x - 5}{5x}$$
ĐKXĐ: x ≠ -3, x ≠ 0, x ≠ 3
= (x + 9)/[(x - 3)(x + 3)] - 3/[x(x + 3)]
= x(x + 9) - 3(x - 3) / [x(x - 3)(x + 3)]
= x² + 9x - 3x + 9 / [x(x - 3)(x + 3)]
= (x² + 6x + 9) / [x(x - 3)(x + 3)]
= (x + 3)² / [x(x - 3)(x + 3)]
= (x + 3)/[x(x - 3)](x - 12)/(6x - 36) + 6/(x² - 6x)
ĐKXĐ: x ≠ 0, x ≠ 6
= (x - 12)/[6(x - 6)] + 6/[x(x - 6)]
= x(x - 12) + 36 / [6x(x - 6)]
= x² - 12x + 36 / [6x(x - 6)]
= (x - 6)² / [6x(x - 6)]
= (x - 6)/(6x)(3x + 5)/(x² - 5x) + (25 - x)/(25 - 5x)
ĐKXĐ: x ≠ 0, x ≠ 5
= (3x + 5)/[x(x - 5)] + (25 - x)/[5(5 - x)]
= (3x + 5)/[x(x - 5)] - (25 - x)/[5(x - 5)]
= 5(3x + 5) - x(25 - x) / [5x(x - 5)]
= 15x + 25 - 25x + x² / [5x(x - 5)]
= x² - 10x + 25 / [5x(x - 5)]
= (x - 5)² / [5x(x - 5)]
= (x - 5)/(5x)