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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) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
(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: \(=x^2+4x+3+11\)
\(=x^2+4x+14\)
\(=x^2+4x+4+10=\left(x+2\right)^2+10>=10\)
Dấu '=' xảy ra khi x=-2
b: \(-4x^2+4x+5\)
\(=-\left(4x^2-4x-5\right)\)
\(=-\left(4x^2-4x+1-6\right)\)
\(=-\left(2x-1\right)^2+6< =6\)
Dấu '=' xảy ra khi x=1/2
c: \(-x^2+6x-4\)
\(=-\left(x^2-6x+4\right)\)
\(=-\left(x^2-6x+9-5\right)\)
\(=-\left(x-3\right)^2+5< =5\)
Dấu '=' xảy ra khi x=3
giúp mik nhanh vs nha
\(x^2+4xy+2y^2-22y+173\)
\(=\left(x^2+4xy+4y^2\right)-2\left(y^2+11y+\dfrac{121}{4}\right)+\dfrac{467}{2}\)\(=\left(x+2y\right)^2-2\left(y+\dfrac{11}{2}\right)^2+\dfrac{467}{2}\ge\dfrac{467}{2}\forall x;y\)Vậy GTNN của biểu thức là; \(\dfrac{467}{2}\) khi \(\left\{{}\begin{matrix}x+2y=0\\y+\dfrac{11}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-11=0\\y=-\dfrac{11}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=-\dfrac{11}{2}\end{matrix}\right.\)Học tốt nha<3
Câu h đề không đẹp lắm, sửa thành-2x nha
f) x2-2x+5
=x2-2x+1+4
=(x-1)2+4
Vì: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)
Min = 4 khi x=1
g) 2x2-6x
= \(\sqrt{2x}^2-2.\sqrt{2x}.\dfrac{3\sqrt{2}}{2}+\left(\dfrac{3\sqrt{2}}{2}\right)^2-\left(\dfrac{3\sqrt{2}}{2}\right)^2\)
= \(\left(\sqrt{2x}-\dfrac{3\sqrt{2}}{2}\right)^2-\dfrac{9}{2}\)
Tương tự bài trên
h) x2+y2-2x+6y+10
=(x2-2x+1)+(y2+6y+9)
=(x-1)2+(y+3)2
Min=0 khi x=1; y=-3
nói thật bn xạo lz vc đề thế nào thì để đó chứ ko đẹp thì nó ko có Min à
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.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
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) \(E=4x^2+y^2-4x-2y+3=\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của E là 1 khi \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)vậy GTNN của E là 1 khi \(x=\dfrac{1}{2};y=1\)
b) \(G=x^2+2y^2+2xy-2y=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1\)
\(=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của G là \(-1\) khi \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\) vậy GTNN của G là \(-1\) khi \(y=1;x=-1\)
c) \(H=x^2+14x+y^2-2y+7=\left(x^2+14x+49\right)+\left(y^2-2y+1\right)-43\)
\(=\left(x+7\right)^2+\left(y-1\right)^2-43\ge-43\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của H là \(-43\) khi \(\left\{{}\begin{matrix}\left(x+7\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=1\end{matrix}\right.\) vậy GTNN của H là \(-43\) khi \(x=-7;y=1\)
d) câu này hình như đề sai