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a/\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow4x^4+16x^3+23x^2+14x-15=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)\left(x^2+2x+3\right)=0\)
Tới đây thì đơn giản rồi b tự làm nhé
b/ \(3x^4-13x^3+16x^2-13x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-1\right)\left(x^2-x+1\right)=0\)
Tới đây thì bạn làm tiếp nhé
c/ \(\left(x+3\right)^4+\left(x+5\right)^4=16\)
\(\Leftrightarrow2x^4+32x^3+204x^2+608x+690=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(x^2+8x+23\right)=0\)
Bạn làm tiếp nhé
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
b) \(\left(2x-3\right)^2=\left(x-5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-3+x-5\right)\left(2x-3-x+5\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{8}{3}\end{array}\right.\)
c) \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x\right)^2-3^2=0\)
\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-\frac{1}{8}\)
Vậy.........
c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x=-255\)
\(\Leftrightarrow x=-127,5\)
Vậy.............
có j sai xót mong m.n bỏ qua☺
a) \(25x^2-9=0\)
<=> \(\left(5x\right)^2=9\)
<=> \(\left(5x\right)^2=3^2\)
<=> \(5x=3\)
<=> \(x=\frac{3}{5}\)
b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)
<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)
<=> \(x^2+8x+16-x^2+1=16\)
<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)
<=> \(8x+17=16\)
<=> \(8x=-1\)
<=> \(x=\frac{-1}{8}\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)
<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)
<=> \(2x+245=0\)
<=> \(2x=-245\)
<=> \(x=\frac{-245}{2}\)
23: \(=\left(2a-b\right)^2-\left(2a-2b\right)^2\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
24: \(=\left(3a+3b\right)^2-\left(2a-4b\right)^2\)
\(=\left(3a+3b-2a+4b\right)\left(3a+3b+2a-4b\right)\)
\(=\left(a+7b\right)\left(5a-b\right)\)
25: \(=\left(4a-2b\right)^2-\left(4a-4b\right)^2\)
\(=\left(4a-2b-4a+4b\right)\left(4a-2b+4a-4b\right)\)
\(=2b\left(8a-6b\right)\)
=4b(4a-3b)
a,\((x+4)^2-(x+1)(x-1)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)
b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)
\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)
\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)
\(\Rightarrow -x^2-10x-244=0\)
\(\Rightarrow -(x^2-10x+25)-219=0\)
\(\Rightarrow -(x-5)^2-219=0\)
\(\Rightarrow (x-5)^2+219=0\)
Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm
1. \(f\left(x\right)=25x^2-20x+\dfrac{9}{2}\)
=>\(f\left(x\right)=25x^2-20x+4+\dfrac{1}{2}\)
=> \(f\left(x\right)=(25x^2-20x+4)+\dfrac{1}{2}\)
=> \(f\left(x\right)=(5x-2)^2+\dfrac{1}{2}\)
Ta thấy: \((5x-2)^2\ge0\)
=>\(f\left(x\right)=(5x-2)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\)(đpcm)
2. \(f\left(x\right)=4x^2-28x+50\)
=> \(f\left(x\right)=(4x^2-28x+49)+1\)
=> \(f\left(x\right)=(2x-7)^2+1\)
Ta thấy: \((2x-7)^2\ge0\)
=> \(f\left(x\right)=(2x-7)^2+1\ge1>0\) (đpcm)
3. \(f\left(x\right)=-16x^2+72x-82\)
=> \(f\left(x\right)=-(16x^2-72x+82)\)
=> \(f\left(x\right)=-(16x^2-72x+81+1)\)
=> \(f\left(x\right)=-[(4x-9)^2+1]\)
Ta thấy: \((4x-9)^2\ge0\)
=> \((4x-9)^2+1\ge1>0\)
=> \(f\left(x\right)=-[(4x-9)^2+1]< 0\)
5. \(f\left(x;y\right)=4x^2+9y^2-12x+6y+11\)
=> \(f\left(x;y\right)=4x^2+9y^2-12x+6y+9+1+1\)
=> \(f\left(x;y\right)=(4x^2-12x+9)+(9y^2+6y+1)+1\)
=> \(f\left(x;y\right)=(2x-3)^2+(3y+1)^2+1\)
Ta thấy: \((2x-3)^2\ge0\)
\((3y+1)^2\ge0\)
=> \(f\left(x;y\right)=(2x-3)^2+(3y+1)^2+1\) \(\ge1>0\) (đpcm)