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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
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)
a) \(\left(x^2-3x+1\right)\left(x^2+5x+1\right)=2x^2\)
\(\Rightarrow\)Cậu xem lại đề xem có sai chỗ nào không nhé !
b) \(x^4-9x\left(x^2-2\right)+16x^2+4=0\)
\(\Leftrightarrow x^4-9x^3+18x+16x^2+4=0\)
\(\Leftrightarrow x^4-4x^3-2x^2-5x^3+20x^2+10x-2x^2+8x+4=0\)
\(\Leftrightarrow x^2\left(x^2-4x-2\right)-5x\left(x^2-4x-2\right)-2\left(x^2-4x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x-2\right)\left(x^2-5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4x-2=0\\x^2-5x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\pm\sqrt{6}\\x=\frac{5\pm\sqrt{33}}{2}\end{cases}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{2\pm\sqrt{6};\frac{5+\sqrt{33}}{2}\right\}\)
b) \(ĐKXĐ:x\ne1;x\ne\frac{2}{3}\)
\(\frac{2x}{3x^2-5x+2}+\frac{13x}{3x^2+x+2}=0\)
\(\Leftrightarrow\frac{2x\left(3x^2+x+2\right)+13x\left(3x^2-5x+2\right)}{\left(3x^2-5x+2\right)\left(3x^2+x+2\right)}=0\)
\(\Leftrightarrow\frac{6x^3+2x^2+4x+39x^3-65x^2+26x}{\left(3x^2-5x+2\right)\left(3x^2+x+2\right)}=0\)
\(\Leftrightarrow45x^3-63x^2+30x=0\)
\(\Leftrightarrow3x\left(15x^2-21x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\15x^2-21x+10=0\left(ktm\right)\end{cases}}\)
Vậy x = 0 là nghiệm của phương trình.
\(\frac{x^2-x-6}{x-3}=\frac{x^2-3x+2x-6}{x-3}=\frac{x\left(x-3\right)+2\left(x-3\right)}{\left(x-3\right)}=x+2=0\Leftrightarrow x=-2\)
\(\frac{x^2+2x-\left(3x+6\right)}{x+2}=\frac{x\left(x+2\right)-3\left(x+2\right)}{x+2}=x-3=0\Leftrightarrow x=3\)
\(\frac{4}{x-2}-\left(x-2\right)=0\Leftrightarrow\frac{4}{a}-a=0\left(a=x-2\right)\Leftrightarrow\frac{4}{a}=a\Leftrightarrow a^2=4\Leftrightarrow a=\pm2\Leftrightarrow x=4\text{ hoặc 0}\)
a) ĐKXĐ: x \(\ne\)3
Ta có: \(\frac{x^2-x-6}{x-3}=0\)
<=> x2 - x - 6 = 0
<=> x2 - 3x + 2x - 6 = 0
<=> (x + 2)(x - 3) = 0
<=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=3\left(vn\right)\end{cases}}\)
Vậy S = {-2}
b) ĐKXĐ: x \(\ne\)-2
Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x+2}=0\)
<=> \(x\left(x+2\right)-3\left(x+2\right)=0\)
<=> \(\left(x-3\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=-2\left(vn\right)\end{cases}}\)
Vậy S = {3}
c) ĐKXĐ: x \(\ne\)2
Ta có: \(\frac{4}{x-2}-x+2=0\)
<=> \(\frac{4-\left(x-2\right)^2}{x-2}=0\)
<=> \(\left(2-x+2\right)\left(2+x-2\right)=0\)
<=> \(x\left(4-x\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\4-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy S = {0; 4}
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
a)3x2+4x-9x-12=0
=>(3x2+4x)-(9x+12)=0
=> x(3x+4)-3(3x+4)=0
=> (x-3)(3x+4)=0 =>x-3=0 hoặc 3x+4=0
=>tự tính
b)7x2-9x+2=0
=>7x2-7x-2x+2=0
=>(7x2-7x)-(2x-2)=0
=>7x(x-1)-2(x-1)=0
=>(7x-2)(x-1)=0
=>như câu a
bạn chỉ biết làm 2 câu thôi
a) \(3x^2-5x-12=0\)
\(\Leftrightarrow3x^2+4x-9x-12=0\)
\(\Leftrightarrow x\left(3x+4\right)-3\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=3\end{cases}}\)
b) \(7x^2-9x+2=0\)
\(\Leftrightarrow7x^2-7x-2x+2=0\)
\(\Leftrightarrow7x\left(x-1\right)-2\left(x-1\right)=0\).
\(\Leftrightarrow\left(7x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=1\end{cases}}\)