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\(1,3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
\(2,5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(5x-1\right)\left(x-3\right)\)
\(3,5x\left(x-3\right)-x+3\)
\(=\left(5x-1\right)\left(x-3\right)\)
\(4,x^2-4+3y\left(x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)+3y\left(x+2\right)\)
\(=\left(x+2\right)\left(x-2+3y\right)\)
1) 3x(x-2) - 2(x-2)=(x-2)(3x-2)
2) 5x(x-3) +x-3=5x(x-3)+(x-3)=(x-3)(5x+1)
3) 5x(x-3) - x+3=5x(x-3) -(x-3)=(x-3)(5x-1)
4) x^2 -4+3y(x+2)=(x^2-4)+3y(x+2)=(x-2)(x+2)+3y(x+2)=(x+2)(x-2+3y)
Hok tốt nha!!!=.=
a) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 4
<=> x3 - 9x2 + 27x - 27 - ( x3 - 27 ) + 9( x2 + 2x + 1 ) = 4
<=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 = 4
<=> 45x + 9 = 4
<=> 45x = -5
<=> x = -5/45 = -1/9
b) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 17
<=> x( x2 - 25 ) - ( x3 + 8 ) = 17
<=> x3 - 25x - x3 - 8 = 17
<=> -25x - 8 = 17
<=> -25x = 25
<=> x = -1
1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x=t\)
\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)
\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)
Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)
\(\Delta=7^2-4.11=5\)
\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)
2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x=t\)
\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)
\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)
Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)
\(\Delta=\left(-8\right)^2-4.11=20\)
\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)
Ta có: A = x2 + y2 - 2(x + y) + 5
A = x2 + y2 - 2x - 2y + 5
A = (x2 - 2x +1) + (y2 - 2y + 1) + 3
A = (x - 1)2 + (y - 1)2 + 3
Do (x - 1)2 \(\ge\)0 \(\forall\)x; (y - 1)2 \(\ge\)0 \(\forall\)y
=> (x - 1)2 + (y - 1)2 + 3 \(\ge\)3 > 0 \(\forall\)x;y
=> A > 0 \(\forall\)x; y
Ta có: A = (x + 1)(x + 2)(x + 3)(x + 4) - 24
A = (x + 1)(x + 4)(x + 2)(x + 3) - 24
A = (x2 + 5x + 4)(x2 + 5x + 6) - 24
Đặt x2 + 5x + 4 = k
=> k(k + 2) - 24 = k2 + 2k - 24 = k2 + 6k - 4k - 24 = k(k + 6) - 4(k + 6) = (k - 4)(k + 6)
=> (x2 + 5x + 4 - 4)(x2 + 5x + 4 + 6) = (x2 + 5x)(x2 + 5x + 10) = x(x + 5)(x2 + 5x + 10)
Do x + 5 \(⋮\)x + 5 => x(x + 5)(x2 + 5x + 10) \(⋮\)x + 5
thế cần CM cho x khác 5 ko