Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) x2-4x+5+y2+2y=0
<=>x2-4x+4+y2+2y+1=0
<=>(x-2)2+(x+1)2=0
<=>x-2=0 và x+1=0
<=>x=2 và x=-1
2)2p.p2-(p3-1)+(p+3)2p2-3p5
<=>2p3-p3+1+2p3+6p2-3p5
<=>3p3+6p2-3p5+1
3)(0.2a3)2-0.01a4(4a2-100)=0,04a6-0,04a6+1
=1
4)a) x(2x+1)-x2(x+20)+(x3-x+3)=2x2+x-x3-20x2+x3-x+3
=-18x2+3(đề sai)
b) x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2)=3x3-x2+5x-2x3-3x+16-x3+x2-2x
=16
Vậy x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2) không phụ thuộc vào x
5)a) x(y-z)+y(z-x)+z(x-y)=xy-xz+yz-xy+xz-yz=0
b) x(y+z-yz)-y(z+x-xz)+z(y-x)=xy+xz-xyz-yz-xy+xyz+yz-xz=0
6)M+(12x4-15x2y+2xy2+7)=0
<=>M =-(12x4-15x2y+2xy2+7)
<=>M =-12x4+15x2y-2xy2-7
\(A=-x^2-4x-2\)
\(\Leftrightarrow-A=x^2+4x+2\)
\(\Leftrightarrow-A=x^2+4x+4-2\)
\(\Leftrightarrow-A=\left(x+2\right)^2-2\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2-2\ge-2\)hay \(-A\ge-2\)
\(\Rightarrow A\le2\)
Vậy GTLN của A là 2\(\Leftrightarrow x=-2\)
a) x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2 ( x + 1 ) + x ( x + 1 )
= ( x2 + x ) ( x + 1 )
\(A=4x^2-2\left(y+2,5x^2\right)+x^2-4y\)
\(=4x^2-2y-5x^2+x^2-4y=-6y\)
\(B=\left(x+y\right).\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)-\left(x^5+y^5-8\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5-x^5-y^5+8\)
\(=8\)
Vậy BT B ko phụ thuộc vào biến
câu sau tương tự
\(5x\left(x+1\right)-3\left(x-5\right)+4\left(3x-6\right)=2x^2-7\)
\(\Rightarrow5x^2+5x-3x+15+12x-24=2x^2-7\)
\(\Rightarrow5x^2+14x-9=2x^2-7\Rightarrow5x^2+14x-9-2x^2+7=0\)
\(\Rightarrow3x^2+14x-2=0\)
\(\Rightarrow3\left(x^2+\frac{14}{3}x-\frac{2}{3}\right)=0\Rightarrow x^2+2.x.\frac{7}{3}+\frac{49}{9}-\frac{55}{9}=0\)
\(\Rightarrow\left(x+\frac{7}{3}\right)^2=\frac{55}{9}\Rightarrow x+\frac{7}{3}\in\left\{\sqrt{\frac{55}{9}};-\sqrt{\frac{55}{9}}\right\}\Rightarrow x\in\left\{\sqrt{\frac{55}{9}}-\frac{7}{3};-\sqrt{\frac{55}{9}}-\frac{7}{3}\right\}\)
a) \(x^2+2x+9=\left(x^2+2x+1\right)+8=\left(x+1\right)^2+8\)
Ta có :
\(\left(x+1\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2+8\ge8>0\)
Do đó đa thức vô nghiệm.
Vậy...
b) \(y^2-y+1=\left(y^2-2.\frac{1}{2}y+\frac{1}{4}\right)+\frac{3}{4}=\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có :
\(\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
Do đó đa thức vô nghiệm.
Vậy ...
c) \(2y^2-2y+4\)
\(=2y^2-2y+\frac{1}{2}+\frac{7}{2}\)
\(=2\left(y^2-2.\frac{1}{2}.y+\frac{1}{4}\right)+\frac{7}{2}\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)
Ta có :
\(\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\ge\frac{7}{2}>0\)
Do đó đa thức vô nghiệm
Vậy...
d) \(3x^4+x^2+2\)
\(=2x^4+\left(x^4+2.\frac{1}{2}x^2+\frac{1}{4}\right)+3\)
\(=2\left(x^2\right)^2+\left(x^2+\frac{1}{2}\right)^2+3\)
Ta có :
\(\left(x^2\right)^2\ge0\)
\(\Rightarrow2\left(x^2\right)^2\ge0\)
\(\left(x^2+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(x^2\right)^2+\left(x^2+\frac{1}{2}\right)^2+3\ge3>0\)
Do đó đa thức vô nghiệm.
Vậy ...
e) \(x^2+x+1=\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có :
\(\left(x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
Do đó đa thức vô nghiệm.
Vậy ...
f) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-5\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=5\\x=1\end{cases}.}\)
g) \(x^3-x^2+2\)
\(=x^3-x^2+2x-2x+2\)
\(=\left(x^3-x\right)-\left(x^2-x\right)-2\left(x-1\right)\)
\(=x\left(x^2-1\right)-x\left(x-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-x\left(x-1\right)-2\left(x-1\right)\)
\(=\left[x\left(x+1\right)-x-2\right]\left(x-1\right)\)
\(=\left(x^2+x-x-2\right)\left(x-1\right)\)
\(=\left(x^2-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-2=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\in\left\{-\sqrt{2};\sqrt{2}\right\}\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x\in\left\{-\sqrt{2};\sqrt{2}\right\}\\x=1\end{cases}}.\)
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
Th1: 2x+3 ≥ 0
Khi đó: |2x+3| =x+2
(2x+3)= x+2
- 2x+3= x+2
-2x-x= 2-3
x= -1
Th2: 2x+3 < 0
Khi đó: |2x+3|=x+2
-(2x+3) = x +2
-2x-3 = x+2
-3x = 5
x=-5/3
Vậy x= -1
x= -5/3
Lớp 6 cugx học dạng v nè
Câu b nha