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a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1-x+3\right)^2=\left(x+4\right)^2\)
b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)
c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)
d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y+z\right)\left(x+y\right)\)
f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
1,
a, (2x + 1- x + 3)2 = (x+4)2
b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)
c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)
d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)
e,\(\left(x+y\right)^2-z\left(x+y\right)\)=\(\left(x+y\right)\left(x+y-z\right)\)
f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)
\(=\left(5+2x+y\right)\left(5-2x-y\right)\)
Chúc các bn hc tốt
1: \(=a\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(a-4\right)\)
2: \(=x\left(x+b\right)+a\left(x+b\right)=\left(x+b\right)\left(x+q\right)\)
3: \(=a\left(x+1\right)-b\left(x+1\right)+c\left(x+1\right)\)
\(=\left(x+1\right)\left(a-b+c\right)\)
6: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
B3) a) x(x-5)-4(x-5)=0
<=> (x-4)(x-5)=0
TH1 :x-4=0
<=.x=4
TH2 : x-5=0
<=>x=5
b) x(x-6)-7x-42=0
<=>x(x+6)-7(x+6)=0
<=>(x-7)(x+6)=0
th1;x-7=0
<=>x=7
th2; x+6=0
<=>x=-6
c)x^3-5x^2+x-5=0
<=> x(x^2+1)-5(x^2+1)=0
<=> (x-5)(x^2+1)=0
th1:x-5=0
<=>x=5
TH2 : x^2+1=0
<=> x^2=-1 ( vo li )
=> th2 ko tồn tại
nho thick nha
Bài 3
a, x(x-5)-4(x-5)=0
(x-4)(x-5)=0
=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
b,x(x+6)-7(x+6)=0
(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
c,x^2(x-5)+(x-5)=0
(x^2+1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
Bài 2:
a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)
\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)
\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)
b: \(B=-x^2-y^2+xy+2x+2y\)
\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)
\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)
\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)
\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)
Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0
=>y=2 và 2x=y+2=2+2=4
=>x=2 và y=2
Bài 1:
d: \(D=2x^2+3y^2+4xy-8x-2y\)
\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)
\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)
\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)
f: \(F=2x^2+8xy+11y^2-4x-2y+6\)
\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)
\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)
\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi y+1=0 và x+2y-1=0
=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3
h: \(H=x^2+y^2-xy-x+y+1\)
\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)
\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)
\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)
\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)
Bài 2:
a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)
\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)
\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)
b: \(B=-x^2-y^2+xy+2x+2y\)
\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)
\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)
\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)
\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)
Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0
=>y=2 và 2x=y+2=2+2=4
=>x=2 và y=2
Bài 1:
d: \(D=2x^2+3y^2+4xy-8x-2y\)
\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)
\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)
\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)
f: \(F=2x^2+8xy+11y^2-4x-2y+6\)
\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)
\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)
\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi y+1=0 và x+2y-1=0
=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3
h: \(H=x^2+y^2-xy-x+y+1\)
\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)
\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)
\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)
\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
ai ,mình tích lại
2x^2+xy+2y^2 = 5/4.(x+y)^2 + 3/4. (x-y)^2 >= 5/4. (x+y)^2
=> cbh(2x^2+xy+2y^2) >= cbh5 / 2. (x+y)
tương tự với 2 căn còn lại.. cộng vế ta có VT >= cbh5 ( x+y+z) = cbh5 : dpcm
dau = cay ra <=> x=y=z=1/3