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Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
\(2x^2+6x-8=0\)
<=> \(2x^2-2x+8x-8=0\)
<=> \(2x\left(x-1\right)+8\left(x-1\right)=0\)
<=> \(\left(2x+8\right)\left(x-1\right)=0\)
<=> \(\hept{\begin{cases}2x+8=0\\x-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-4\\x=1\end{cases}}\)
\(2x^2-x-1=0\)
<=> \(2x^2-2x+x-1=0\)
<=> \(2x\left(x-1\right)+\left(x-1\right)=0\)
<=> \(\left(2x+1\right)\left(x-1\right)=0\)
<=> \(\hept{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
\(4x^2-5x-9=0\)
<=> \(4x^2+4x-9x-9=0\)
<=> \(4x\left(x+1\right)-9\left(x+1\right)=0\)
<=> \(\left(4x-9\right)\left(x+1\right)=0\)
<=> \(\hept{\begin{cases}4x-9=0\\x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)
học tốt
\(2x^2+6x-8=0\)
\(< =>2x^2-2x+8x-8=0\)
\(\Leftrightarrow2x\left(x-1\right)+8\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x+8=0\)hoặc \(x-1=0\)
\(\Leftrightarrow x=-4\)hoặc \(x=1\)
2(x + 7) - (2x + 3).(x - 1) - 8 = 6x
<=> (2x + 14) - (2x + 3)(x - 1) - 8 - 6x = 0
<=> 2x + 14 - (2x2 + 3x - 2x - 3) - 8 - 6x = 0
<=> 2x + 14 - (2x2 + x - 3) - 8 - 6x = 0
<=> 2x + 14 - 2x2 - x + 3 - 8 - 6x = 0
<=> -2x2 - 5x + 6 = 0
<=> 2x2 + 5x - 6 = 0
<=> \(x^2+\frac{5}{2}x-3=0\)
\(\Leftrightarrow x^2+2.x.\frac{5}{4}+\frac{25}{16}-\frac{73}{16}=0\)
\(\Leftrightarrow x^2+2.x.\frac{5}{4}+\frac{25}{16}=\frac{73}{16}\)
\(\Leftrightarrow\left(x+\frac{5}{4}\right)^2=\frac{73}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{4}=\frac{73}{16}\\x+\frac{5}{4}=-\frac{73}{16}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{73}{16}-\frac{5}{4}\\x=-\frac{73}{16}-\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{53}{16}\\x=-\frac{93}{16}\end{cases}}\)
2(x+7)-(2x+3)(x-1)-8=6x
\(\Leftrightarrow2x+14-2x^2+2x-3x+3-8=6x\)
\(\Leftrightarrow\) \(-2x^2+2x+2x-3x+3-8+14=6x\)
\(\Leftrightarrow-2x^2+x+9=6x\)
\(\Leftrightarrow-2x^2-5x+9=0\)
\(\Leftrightarrow\left(x-\left(\frac{-5+\sqrt{97}}{4}\right)\right)\left(x+\left(\frac{-5-\sqrt{97}}{4}\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{-5+\sqrt{97}}{4}=0\\x+\frac{-5-\sqrt{97}}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+97}{4}\\x=\frac{5-\sqrt{97}}{4}\end{cases}}\)
nhân đa thức trước bạn nhé!
<=>6x^2 -(6x^2 +4x -9x -6)-1=0
phía trước là dấu trừ nên đổi dấu hạng tử bên trong
<=> 6x^2 -6x^2 -4x +9x+6 -1=0
<=>5x =-5
<=> x=-1
\(6x^2-\left(2x-3\right).\left(3x+2\right)-1=0\) \(0\)
\(< =>6x^2+\left(-2x+3\right).\left(3x+2\right)-1=0\)
\(< =>6x^2-6x^2-4x+9x+6-1=0\)
\(< =>5x=-5\)
\(< =>x=-1\)
a: Ta có: \(2x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
B = x2y2+2x2+24xy+16x+191 = [ (xy)^2 + 24xy + 144] + \(\left[\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.4\sqrt{2}+32\right]\)+15
= (xy+12)^2 +(\(\sqrt{2}x\)+\(4\sqrt{2}\))^2 + 15
( ở đây mik làm tắt) => Min B = 15 khi \(\sqrt{2}x+4\sqrt{2}=0=>x=-4\)và xy+12 = 0 => -4y = -12= > y=3
A= 2x^2+9y^2-6xy-6x-12y+2004
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2004
A = [(x -3y)^2 +4(x -3y) + 4] + (x^2 -10x +25) + 1975
A= (x -3y +2)^2 + (x -5)^2 + 1975
( mik rút mấy cái bước (x-3y+2)^2 = 0, bn làm thì nên thêm vào=> Min A = 1975 vs x= 5 và y = 7/3
D=-x^2+2xy-4y^2+2x+10y-8
D = (-x^2 - y^2 - 1 + 2xy + 2x - 2y) + (-3y^2 + 12y - 12) + 5
D = -(x^2+y^2+1 - 2xy - 2x + 2y) - 3(y^2 - 4y + 4) + 5
D= - (x - y - 1)^2 - 3(y - 2)^2 +5
=> Max D = 5 khi x= 3 và y=2
