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\(B=\left(x^2-8x\right)\left(x^2-8x+24\right)\)
Đặt \(x^2-8x+12=t\) ta có:
\(B=\left(t-12\right)\left(t+12\right)=t^2-144\ge-144\)
Dấu "=" xảy ra khi \(t^2=0\Leftrightarrow t=0\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\Leftrightarrow x=2;x=6\)
\(C=5x^2+9y^2-6xy-12x+13\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-12x+9\right)+4\)
\(=\left(x-3y\right)^2+\left(2x-3\right)^2+4\ge4\)
Dấu "=" xảy ra tại \(x=\frac{3}{2};y=\frac{1}{2}\)
Vì \(a^2+b^2+c^2=1\)
\(\Rightarrow-1\le a,b,c\le1\)
\(\Rightarrow a-1\le0;b-1\le0;c-1\le0\)
Lây cai xau trừ cai trươc được
\(\left(a^3+b^3+c^3\right)-\left(a^2+b^2+c^2\right)=0\)
\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)
Ta co \(VT\le0\)
Dâu = xảy ra khi: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)
\(\Rightarrow S=1\)
4) (3x-2)(x-3)= 3x(x-3)-2(x-3)
=3x.x+3x.(-3)-2.x-2.(-3)
=\(3x^2\)-9x-4x+6
=\(3x^2\)+(-9x-4x)+6
=\(3x^2\)-13x+6
5) (2x+1)(x+3)=2x(x+3)+1(x+3)
=2x.x+2x.3+1.x+1.3
=\(2x^2\)+6x+1x+3
=\(2x^2\)+(6x+1x)+3
=\(2x^2\)+7x+3
6) (x-3)(3x-1)=x(3x-1)-3(3x-1)
=x.3x+x.(-1)-3.3x-3.(-1)
=\(3x^2\)-1x-9x+3
=\(3x^2\)+(-1x-9x)+3
=\(3x^2\)-10x+3
rút gọn biểu thức
A) \(x^2\)-(x+4)(x-1)=\(x^2\)- x(x-1)-4(x-1)
=\(x^2\)-x.x-x.(-1)-4.x-4.(-1)
=\(x^2\)-\(x^2\)+1x-4x+4
=(\(x^2-x^2\))+(1x-4x)+4
= -3x+4
B) x(x+2)-(x-2)(x+4)=x.x+x.2-x(x+4)+2(x+4)
=\(x^2+2x\)-x.x-x.4+2.x+2.4
=\(x^2+2x-x^2-4x+2x+8\)
=(\(x^2-x^2\))+(2x-4x+2x)+8
=8
tính giá trị biểu thức
A=3(x-2)-(2+x)(x-3)
=3.x+3.(-2)-2(x-3)-x(x-3)
=3x-6-2.x-2.(-3)-x.x-x(-3)
=3x-6-2x+6-\(x^2\)+3x
=(3x-2x+3x)+(-6+6)\(-x^2\)
=4x - \(x^2\)
thay x=-8 vào biểu thức thu gọn ta được:
4.(-8)- (-8)\(^2\)
= - 32 +64
= 32
B= x(3-x)-(1+x)(1-x)
=x.3+x.(-x)-1(1-x)-x(1-x)
=3x -\(x^2\)-1.1-1 .(-x)-x.1-x.(-x)
=3x\(-x^2\)-\(1^2\)+1x-1x+\(x^2\)
=(3x+1x-1x)+(\(-x^2+x^2\))-1
=3x-1
thay x=-5 vào biểu thức thu gọn ta được:
3.(-5)-1
=-15-1
=-16
Thu gọn biểu thức
4) (3x - 2) (x - 3)
= ( 3x2 - 2x ) - ( 3x x 3 - 2 x 3 )
= 3x2 - 2x - 3x x 3 + 2 x 3
= 3x2 - 2x - 9x + 6
= 3x2 - 11x + 6
5) (2x + 1) (x + 3)
= ( 2x2 + 1x ) + ( 6x + 3 )
= 2x2 + 1x + 6x + 3
= 2x2 + 7x + 3
6) (x - 3) (3x - 1)
= ( 3x2 - 9x ) - ( x - 3 )
= 3x2 - 9x - x + 3
= 3x2 - 10 + 3
Rút gọn biểu thức
A) x^2 - (x + 4) (x - 1)
= x2 - ( x2 + 4x ) - ( x + 4 )
= x2 - x2 - 4x - x - 4
= -5x - 4
B) x (x + 2) - (x - 2) (x + 4)
= x2 + 2x - ( x2 - 2x ) + ( 4x - 8 )
= x2 + 2x - x2 + 2x + 4x - 8
= 8x - 8
Tính giá trị biểu thức
A = 3 (x - 2) - (2 + x) (x - 3) tại x = - 8
Thế x = -8 vào, ta có :
= 3 ( -8 -2 ) - ( 2 + -8 ) ( -8 - 3 )
= 3 x ( -10 ) - ( - 6 ) ( -11 )
= -30 - 66
= -96
B = x (3 - x) - (1 + x) ( 1 - x) tại x = - 5
Thế x = - 5 vào, ta có :
= -5 ( 3 - -5 ) - ( 1+ -5 ) ( 1 - -5 )
= -5 x 8 - (-4) x 6
= - 40 - -24
= -40 + 24
= -16
100% đúng
hok tốt nha
B1:
a,\(\left(3x-2\right)\left(x-3\right)=3x^2-9x-2x+6=3x^2-11x+6\)
b,\(\left(2x+1\right)\left(x+3\right)=2x^2+6x+x+3=2x^2+7x+3\)
c,\(\left(x-3\right)\left(3x-1\right)=3x^2-x-9x+3=3x^2-10x+3\)
B2:
1)\(x^2-\left(x+4\right)\left(x-1\right)=x^2-\left(x^2-x+4x-4\right)=x^2-x^2+x-4x+4=-3x+4\)
2)\(x\left(x+2\right)-\left(x-2\right)\left(x+4\right)=x^2+2x-\left(x^2+4x-2x-8\right)\)
\(=x^2+2x-x^2-4x+2x+8=8\)
Bài 1 :
1) 4x2 - y2 = ( 2x + y ) ( 2x - y )
2) 9x2 - 4y2 = ( 3x - 2y ) ( 3x + 2y )
3) 4x2 + y2 + 4xy = ( 2x + y )2
Bài 2:
1) 2x2 + 8x = 0
=> 2x ( x + 4 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
2) 3 ( x - 4 ) + x2 - 4x = 0
=> 3 ( x - 4 ) + x ( x - 4 ) = 0
=> ( x - 4 ) ( 3 + x ) = 0
=> \(\orbr{\begin{cases}x-4=0\\3+x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
3) 3 ( x - 2 ) = x2 - 2x
=> 3 ( x - 2 ) - x2 + 2x = 0
=> 3 ( x - 2 ) - x ( x - 2 ) = 0
=> ( x - 2 ) ( 3 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
4) x ( x - 2 ) - 6 ( 2 - x ) = 0
=> x ( x - 2 ) + 6 ( x - 2 ) = 0
=> ( x - 2 ) ( x + 6 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
5) 2x ( x + 5 ) = x2 + 5x
=> 2x ( x + 5 ) - x2 - 5x = 0
=> 2x ( x + 5 ) - x ( x + 5 ) = 0
=> ( x + 5 ) ( 2x - x ) = 0
=> \(\orbr{\begin{cases}x+5=0\\2x-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
6 ) ( x - 2 )2 - x ( x + 3 ) = 9
=> x2 - 4x + 4 - x2 - 3x = 9
=> - 7x + 4 = 9
=> - 7x = 5
=> x = \(-\frac{5}{7}\)
\(1,4x^2-y^2=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(2,9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)
\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(2,3\left(x-4\right)+x^2-4x=0\)
\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(3,3\left(x-2\right)=x^2-2x\)
\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)
\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
\(4,x\left(x-2\right)-6\left(2-x\right)=0\)
\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
