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a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x\right)^2-5^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(-2\right).\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
a,\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(4x^2-25\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-5-2x-7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(-12\right)=0\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
\(b,2x^3+3x^2+2x+3=0\)
\(\Rightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Rightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-3\\x^2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
\(c,x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right).x^3=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)
\(d,x^2\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
a) \(\left(x+17\right).\left(25-x\right)=0\)
\(\Leftrightarrow x+17=0\)hoặc \(25-x=0\)
Từ \(x+17=0\Rightarrow x=0-17=-17\)
Từ \(25-x=0\Rightarrow x=25-0=25\)
Vậy \(x=-17\) hoặc \(25\)
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x\right)^2-3^2=0\)
\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{3}{5}\end{cases}}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-\frac{1}{8}\)
a) ko hiểu đề bài
b) Ta có (x + 4)2 - (x + 1)(x - 1) = 16
<=> x2 + 8x + 16 - (x2 - 1) = 16
<=> x2 + 8x + 16 - x2 + 1 = 16
<=> 8x + 17 = 16
=> 8x = -1
=> x = \(-\frac{1}{8}\)
1.
a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3-x+3\right)\)
\(=9\left(x-3\right)=9x-27\)
b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)^2=9x^2\)
c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-9\right)-\left(x^4-1\right)\)
\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.