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\(\dfrac{-4+25}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)
= \(\dfrac{-4+25}{x^2-25}-\dfrac{2x^2+x}{x^2-25}+\dfrac{2x\left(x+5\right)}{x^2-25}\)
= \(\dfrac{-4+25-2x^2-x+2x^2+10x}{x^2-25}\)
= \(\dfrac{21+9x}{x^2-25}\)
Ta có: \(\dfrac{-4x^2}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)
\(=\dfrac{-4x^2-2x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x}{x-5}\)
\(=\dfrac{-6x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-6x^2-x+2x^2+10x}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x^2+9x}{\left(x-5\right)\left(x+5\right)}\)
Đặt \(a=24-x,b=x-25\)
Khi đó pt ban đầu trở thành :
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow30a^2+68ab+30b^2=0\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)
Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !
Bài này dễ mà bn
\(\left(10a+5\right)^2=100a^2+100a+25=\left(100a^2+100a\right)100a\left(a+1\right)+25\)
1. (x^2-25)^2-(x-5)^2
=(x-5)2(x+5)2-(x-5)2
=(x-5)2.[(x+5)2-1)
=(x-5)2.(x+5-1)(x+5+1)
=(x-5)2.(x+4)(x+6)
2. (4x^2-25)^2-9(2x-5)^2
=(2x-5)2(2x+5)2-9.(2x-5)2
=(2x-5)2[(2x+5)2-9]
=(2x-5)2(2x+5-3)(2x+5+3)
=(2x-5)2(2x+2)(2x+8)
=4(2x-5)2(x+1)(x+4)
a: \(9x^2-30x+25=0\)
\(\Leftrightarrow3x-5=0\)
hay \(x=\dfrac{5}{3}\)
c: \(9x^2-25=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
a) \(9x^2-30x+25=0\Rightarrow\left(3x-5\right)^2=0\Rightarrow x=\dfrac{5}{3}\)
b) \(25x^2-5x+\dfrac{1}{4}=0\Rightarrow\left(10x-1\right)^2=0\Rightarrow x=\dfrac{1}{10}\)
c) \(9x^2-25=0\Rightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
d) \(\left(2x-1\right)^2-\left(3x+2\right)^2=0\)
\(\Rightarrow\left(2x-1+3x+2\right)\left(2x-1-3x-2\right)=0\)
\(\Rightarrow-\left(5x+1\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)