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a) \(\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
<=> \(9x^2-9x+2=9x^2+6x+1\)
<=> \(15x=1\) <=> \(x=\frac{1}{15}\)
b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)
<=> \(4x^2+3x-1=4x^2-12x+9\)
<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)
c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)
<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)
<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)
d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)
<=> 16 - 9x2 = 12x - 9x2 - 3
<=> 12x = 19
<=> x = 19/12
e) x(x + 1)(x + 2)(x + 3) = 24
<=> (x2 + 3x)(x2 + 3x + 2) = 24
<=> (x2 + 3x)2 + 2(x2 + 3x) - 24 = 0
<=> (x2 + 3x)2 + 6(x2 + 3x) - 4(x2 + 3x) - 24 = 0
<=> (x2 + 3x + 6)(x2 + 3x - 4) = 0
<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)
g) (7x - 2)2 = (7x - 3)(7x + 2)
<=> 49x2 - 28x + 4 = 49x2 - 7x - 6
<=> 21x = 10 <=> x = 10/21
mình chỉ viết đáp án thôi nhé! còn nếu ý nào bạn cần lời giải chi tiết mình sẽ giải cho!
a) S= { -2/3;-3/2}
b) S= {-5;1}
c) S= {-1/2;1}
d) S= {3/7;4}
e) S= {3;5}
NHỚ BẤM ĐÚNG CHO MÌNH NHÉ!
\(a,\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
\(9x^2-3x-6x+2=9x^2+6x+1\)
\(-9x+2-6x-1=0\)
\(-15x+1=0\)
\(-15x=-1\)
\(x=\frac{1}{15}\)
a: \(\Leftrightarrow9x^2-9x+2=9x^2+6x+1\)
=>-3x=-1
hay x=1/3
b: \(\Leftrightarrow4x^2+4x-x-1=4x^2-12x+9\)
=>3x-1=-12x+9
=>15x=10
hay x=2/3
c: \(\Leftrightarrow25x^2+10x+1=25x^2+25x-x-1=24x-1\)
=>10x-24x=-1-1
=>-14x=-2
hay x=1/7
d: \(\Leftrightarrow49x^2-28x+4=49x^2+14x-21x-6\)
=>-28x+4=-7x-6
=>-21x=-10
hay x=10/21
a. \(\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
\(\Leftrightarrow9x^2-9x+2=9x^2+6x+1\)
\(\Leftrightarrow-3x=-1\)
\(\Leftrightarrow x=3\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
â) \(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\left(5-x\right)\left(2+3x\right)=\left(2+3x\right)\left(2-3x\right)\)
\(5-x=2-3x\)
\(2x=-3\)
\(x=\frac{-3}{2}\)
Vậy ......
b) \(25-x^2=4x\left(5+x\right)\)
\(\left(5+x\right)\left(5-x\right)=4x\left(5+x\right)\)
\(5-x=4x\)
\(5x=5\)
x=1
Vậy......
a) \(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
<=> \(\left(5-x\right)\left(2+3x\right)+9x^2-4=0\)
<=> \(\left(5-x\right)\left(2+3x\right)+\left(3x-2\right)\left(3x+2\right)=0\)
<=> \(\left(2+3x\right)\left(3x-2+5-x\right)=0\)
<=> \(\left(2+3x\right)\left(2x+3\right)=0\)
<=> \(\orbr{\begin{cases}2x+3=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}\)
b) \(25-x^2=4x\left(5+x\right)\)
<=> \(25-x^2-4x\left(5+x\right)=0\)
<=> \(\left(5-x\right)\left(5+x\right)-4x\left(5+x\right)=0\)
<=> \(\left(5+x\right)\left(5-x-4x\right)=0\)
<=> \(\left(5+x\right)\left(5-5x\right)=0\)
<=> \(\orbr{\begin{cases}5+x=0\\5-5x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-5\\x=1\end{cases}}\)
a: \(\Leftrightarrow\left(3x+2\right)\left(5-x\right)=-9x^2+4\)
\(\Leftrightarrow\left(3x+2\right)\left(5-x\right)+\left(3x+2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(2x+3\right)=0\)
=>x=-2/3 hoặc x=-3/2
b: \(\Leftrightarrow4x\left(x+5\right)+x^2-25=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x-5\right)=0\)
=>x=-5 hoặc x=1
c: \(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
=>x=1 hoặc x=-1/2