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Bài 2:
a: \(A=-3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)
\(=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Dấu '=' xảy ra khi x=2/3
b: \(B=-x^2+5x+3\)
\(=-\left(x^2-5x-3\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)
Dấu '=' xảy ra khi x=5/2
a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
b: \(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=3 hoặc x=-4
c: \(3x^2+2x-5=0\)
\(\Leftrightarrow3x^2+5x-3x-5=0\)
=>(3x+5)(x-1)=0
=>x=1 hoặc x=-5/3
d: \(x^4-2x^2-3=0\)
\(\Leftrightarrow x^4-3x^2+x^2-3=0\)
\(\Leftrightarrow x^2-3=0\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
Bài 2:
a: \(x^2-16-\left(x+4\right)=0\)
=>(x+4)(x-4)-(x+4)=0
=>(x+4)(x-5)=0
=>x=5 hoặc x=-4
b: \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+1=0\)
=>-6x+2=0
=>-6x=-2
hay x=1/3
c: \(4x^2+9=-12x^2\)
\(\Leftrightarrow4x^2+12x^2=-9\)
\(\Leftrightarrow16x^2=-9\)(vô lý)
Do đó: \(x\in\varnothing\)
d: \(4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
=>x=1 hoặc x=1/4
e: \(4x^2-4x+3=0\)
\(\Leftrightarrow4x^2-4x+1+2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=-2\)(vô lý)
Do đó: \(x\in\varnothing\)
a: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
b: Sửa đề: \(\left(x^3+6x^2+12x+8\right)+3\left(x^2+4x+4\right)+3\left(x+2\right)\)
\(=x^3+6x^2+12x+8+3x^2+12x+12+3x+6\)
\(=x^3+9x^2+27x+26\)
a) Ta có:
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).
Bài 1:
a, \(2x\left(y-z\right)+5y\left(z-y\right)=2x\left(y-z\right)-5y\left(y-z\right)\)
\(=\left(y-z\right)\left(2x-5y\right)\)
b, \(x^3-3x^2+3x-1=x^3-x^2-2x^2+2x+x-1\)
\(=x^2.\left(x-1\right)-2x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)=\left(x-1\right)\left(x^2-x-x+1\right)\)
\(=\left(x-1\right)\left(x-1\right)^2=\left(x-1\right)^3\)
c, \(7x^2-7xy-4x+4y=7x.\left(x-y\right)-4.\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
d, \(x^2-6x+8=x^2-2x-4x+8\)
\(=x.\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)
Chúc bạn học tốt!!!
1)
a) \(2x\left(y-z\right)+5y\left(z-y\right)\)
\(=2x\left(y-z\right)-5y\left(y-z\right)\)
\(=\left(y-z\right)\left(2x-5y\right)\)
b) \(x^3-3x^2+3x-1\)
\(=x^3-3.x^2.1+3.x.1^2-1^3\)
\(=\left(x-1\right)^3\)
c) \(7x^2-7xy-4x+4y\)
\(=7x\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
d) \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
2)
a) \(\left(5x^2+3x-1\right)\left(x+3\right)\)
\(=5x^3+3x^2-x+15x^2+9x-3\)
\(=5x^3+3x^2+15x^2-x+9x-3\)
\(=5x^3+18x^2+8x-3\)
b) \(\left(x^3+2x^2+3x-1\right):\left(x^2-2\right)\)
\(=x+2+\dfrac{5x+3}{x^2-2}\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
= \(\left[\left(x+y+z\right)-\left(x+y\right)\right]^2\)
= \(z^2\)
Ta có:(x + y + z)2 - 2(x + y + z) (x + y) + (x + y)2
=[(x+y+z)-(x+y)]2=z2
(x+1)(x+3) +11
=x^2 +4x+3+11
=x^2 +4x+14
=x^2 +4x+4+10
=(x+2)^2 +10
có : (x+2)^2 \(\ge\)0 \(\forall\)x\(\in\)R
10 > 0 \(\forall\)R
=> (x+2)^2 +10 \(\ge\)10
dấu "=" xảy ra khi và chỉ khi :
(x+2)^2=0 \(\Leftrightarrow\)x+2=0 \(\Leftrightarrow\)x=-2
Vậy (x+1)(x+3)+10 đạt giá trị nhỏ nhất là 10 \(\Leftrightarrow\)x=-2
b, 5-4x^2 +4x
=-(4x^2 -4x-5)
=-(2x-1)^2 +4
có (2x-1)^2 \(\ge\)0\(\forall\) x\(\in\)R=> -(2x-1)^2 \(\le\)0 \(\forall\)x \(\in\)R
4>0 \(\forall\)R
=> -(2x-1)^2+4 \(\le\)4
dấu "=" xảy ra \(\Leftrightarrow\)-(2x-1)^2=0
\(\Leftrightarrow\)2x= 1 <=> x=0,5
vậy 5-4x^2+4x đạt giá trị lớn nhất là 4 <=> x=0,5
nếu đúng thì like nha
a) A= 3.(x2-2xy+y2)- 2. (x2+2xy+y2) - x2-y2
A= 3.x2-2xy+y2-2. x2+2xy+y2-x2-y2