chung minh rang 4x2- 4x+ 10>0 voi moi so thuc x
rut gon
x^2 - 4x+ 4
-2x + x^2
va tinh gia tri cua A khi x = -1/2
b) phan tich da thuc thanh nhan tu
a^3 + 8a- 4a - 8
giup minh nhe
Những câu hỏi liên quan
phan tich da thuc sau thanh nhan tu
xy(x+y) + yz(y+z) + xz(x+z) + 2xyz
tinh gia tri bieu thuc
3(x-3)(x+7) + (x-4)2 + 48 tai x = 0,5
chung minh rang
x2 - 6x + 10 >0 voi moi x
4x - x2 - 5 <0 voi moi x
1, xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= x2y+xy2+y2z+yz2+x2z+xz2+2xyz
=(x2y+x2z+xz2+xyz) + ( xy2+y2z+yz2+xyz)
=x(xy+xz+z2+yz)+y(xy+yz+z2+xz)
=(xy+xz+yz+z2).(x+y)
=(x(y+z)+z(y+z)).(x+y)
=((y+z).(x+z)).(x+y)= (x+y)(x+z)(y+z)
2. 3(x-3)(x-7)+(x-4)2+48
=3(x2+4x-21)+x2-8x+16+48
=4x2-4x+1 = (2x-1)2
Thay x=0,5 vào bt trên, ta có : (2.0,5 -1)2=0
3, x2-6x+10
= x2-2.3.x+9+1
=(x-3)2+1 \(\ge\)1 >0 ( do (x-3)2 >=0 với mọi x)
=> x26x+10 >0 với mọi x
4x-x2-5
=-(x2-4x+5)
=- (x2-2.2x+4+1)
= - ((x-2)2+1) = -(x-2)2-1\(\le\)-1 < 0 ( do (x-2)2\(\ge\)0 với mọi x => - (x-2)2\(\le\)0 với mọi x)
vậy, 4x-x2-5<0 với mọi x
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Ta có : x2 - 6x + 10
= x2 - 6x + 9 + 1
= (x - 3)2 + 1
Mà (x - 3)2 \(\ge0\forall x\)
Nên : (x - 3)2 + 1 \(\ge1\forall x\)
=> (x - 3)2 + 1 \(>0\)(đpcm)
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1) Phan tich da thuc sau thanh nhan tu: x2-x-2008.2009
2) Chung minh rang voi moi x,y,z ta luon co: x2+4y2+z2>=2x+12y+4z
3) Cho a-b=4. Tinh gia tri cua bieu thuc: a3-12ab-b3
cac ban lam duoc cau nao thi giup mik nha. mik dang can gap lam
BAI 1.phan tich cac da thuc sau thanh nhan tu:
a,2x^2-2xy-5x+5y
b,8x^2+4xy-2ax-ay
c,x^3-4x^2+4x
d,2xy-x^2-y^2+16
e,x^2-y^2-2yz-z^2
g,3a^2-6ab+3b^2-12c^2
BAI 2.tinh nhanh
a,37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5
b,35^2+40^2-25^2+80.35
BAI 3. Tim x biet:
a,x^3-1/9x0
b,2x-2y-x^2+2xy-y^20
c,x(x-3)+x-30
d,x^2(x-3)+27-9x0
BAI 4.Phan tich cac da thuc sau thanh nhan tu
a,x^2-4x+3
goi y :tach-4x-x3xhoac tach3-1+4
b,x^2+x-6
c,x^2-5x+6
d,x^4+4 (goi y:them va bot 4x^2)
BAI 5.Chung minh rang;...
Đọc tiếp
BAI 1.phan tich cac da thuc sau thanh nhan tu:
a,2x^2-2xy-5x+5y
b,8x^2+4xy-2ax-ay
c,x^3-4x^2+4x
d,2xy-x^2-y^2+16
e,x^2-y^2-2yz-z^2
g,3a^2-6ab+3b^2-12c^2
BAI 2.tinh nhanh
a,37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5
b,35^2+40^2-25^2+80.35
BAI 3. Tim x biet:
a,x^3-1/9x=0
b,2x-2y-x^2+2xy-y^2=0
c,x(x-3)+x-3=0
d,x^2(x-3)+27-9x=0
BAI 4.Phan tich cac da thuc sau thanh nhan tu
a,x^2-4x+3
goi y :tach-4x=-x3xhoac tach3=-1+4
b,x^2+x-6
c,x^2-5x+6
d,x^4+4 (goi y:them va bot 4x^2)
BAI 5.Chung minh rang;
(3n+4)^2-16 chia het cho 3 voi moi so nguyen n.
BAI 6.Tinh gia tri cua bieu thuc sau:
M=a^3-a^2b-ab^2+b^3 voi a=5,75:b=4,25
BAI 7.Tim x biet:
a,x^2+x=6
b,6x^3+x^2=2x
Bài 1 câu g bạn kia làm sai mình sửa lại nhá
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
\(=3\left[\left(a-b\right)^2-4c^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
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Để mình làm tiếp cho :))
Bài 2 :
Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)
\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)
\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)
\(=37,5.10-7,5.10\)
\(=10.30=300\)
Câu b : \(35^2+40^2-25^2+80.35\)
\(=\left(35^2+80.35+40^2\right)-25^2\)
\(=\left(30+45\right)^2-25^2\)
\(=75^2-25^2\)
\(=\left(75+25\right)\left(75-25\right)\)
\(=100.50=5000\)
Bài 3 :
Câu a : \(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)
Câu b : \(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)
Câu c :
\(x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(x^2\left(x-3\right)+27-9x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)
Bài 4 :
Câu a :
\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=\left(x^2-x\right)-\left(3x-3\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
Câu b :
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
Câu c :
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
Câu d :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
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Bài 1:
a) \(2x^2-2xy-5x+5y\)
\(=\left(2x^2-2xy\right)-\left(5x-5y\right)\)
\(=2x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-5\right)\)
b) \(8x^2+4xy-2ax-ay\)
\(=\left(8x^2+4xy\right)-\left(2ax+ay\right)\)
\(=4x\left(2x+y\right)-a\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x-a\right)\)
c) \(x^3-4x^2+4x\)
\(=x\left(x^2-4x+4\right)\)
\(=x\left(x-2\right)^2\)
d) \(2xy-x^2-y^2+16\)
\(=-\left[\left(x^2-2xy+y^2\right)-16\right]\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)
e) \(x^2-y^2-2yz-z^2\)
\(=-\left[\left(z^2+2yz+y^2\right)-x^2\right]\)
\(=-\left[\left(z+y\right)^2-x^2\right]\)
\(=-\left[\left(z+y+x\right)\left(z+y-x\right)\right]\)
g) \(3a^2-6ab+3b^2-12c^2\)
\(=\left(3a^2-6ab+3b^2\right)-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}\right)^2-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}+\sqrt{12c}\right)\left(\sqrt{3a}+\sqrt{3b}-\sqrt{12c}\right)\)
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Xem thêm câu trả lời
1) Chung minh tong hai so nguyen to la mot hop so
VD: 3+5=8
2) Phan tich da thuc 4x2-a2+y2-16b2+4xy+8ab thanh nhan tu
3) Gia tri nho nhat cua bieu thuc M= (x-1)(x+5)(x2+4x+5) la ...
1/ đề sai vd: 2+3=5 là số nguyên tố
2/ \(4x^2-a^2+y^2-16b^2+4xy+8ab\)
\(=\left[\left(2x\right)^2+2.2xy+y^2\right]-\left[a^2+2.4ab-\left(4b\right)^2\right]\)
\(=\left(2x+y\right)^2-\left(a-4b\right)^2\)
\(=\left(2x+y+a-4b\right)\left(2x+y-a+4b\right)\)
3/
\(M=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+5x-x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-5^2\)
\(=\left(x^2+4x\right)^2-25\)
Vì \(\left(x^2+4x\right)^2\ge0\)
\(\Rightarrow\left(x^2+4x\right)^2-25\ge-25\)
\(\Rightarrow M\ge-25\)
Dấu "=" xảy ra khi x = 0 hoặc x = -4
Vậy Mmin = -25 khi x = 0 hoặc x = -4
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phan tich da thuc sau thanh nhan tu
x^3+3x^2+4x+2
6x^4-x^3-7x^2+x+1
giup minh nhe cam on nhieu
\(a.x^3+3x^2+4x+2\)
\(=x^3+x^2+2x^2+2x+2\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+2\right)\)
\(b.6x^4-x^3-7x^2+x+1\)
\(=6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1\)
\(=6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(6x^3+5x^2-2x-1\right)\)
\(=\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)\)
\(=\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[3x\left(2x-1\right)+\left(2x-1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)\)
k giùm cái cho đỡ buồn!
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cho da thuc f(x)= -2+x^4+2x^2-3x^3+4x^4-5x^4+3x^3+3 chung minh rang da thuc f(x) ko co nghiem tai moi gia tri cua x
cho da thuc f( x) = x4+ 2x3 -x - 2
a, phan tich f(x) thanh nhan tu
b, chung minh f(x) chia het cho 6 voi moi x la so nguyen
a)\(f\left(x\right)=x^4+2x^3-x-2\)
\(=x^4+2x^3+x^2-x^2-x-2\)
\(=\left(x^2+x\right)^2-\left(x^2+x\right)-2\)
Đặt \(x^2+x=t\) ta có:
\(=t^2-t-2\)\(=\left(t-2\right)\left(t+1\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)\)
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Phan tich da thuc thanh nhan tu
1. 2x(2012-x)-2012+x
Chung minh
x2+x+1>0 ( voi moi x)
Chung minh rang bieu thuc sau luon co gia tri duong voi moi gia tri cua x
B=x4-2x3+2x2-4x+5
\(B=x^4-2x^3+2x^2-4x+5\)
\(=\left(x^4-2x^3+x^2\right)+\left(x^2-4x+4\right)+1\)
\(=\left(x^2-x\right)^2+\left(x-2\right)^2+1\)
Vì: \(\begin{cases}\left(x^2-x\right)^2\ge0\\\left(x-2\right)^2\ge0\end{cases}\)\(\Rightarrow\left(x^2-x\right)^2+\left(x-2\right)^2\ge0\)
\(\Rightarrow\left(x^2-x\right)^2+\left(x-2\right)^2+1>0\)
Kết luận...............................................
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B=x4−2x3+2x2−4x+5B=x4−2x3+2x2−4x+5
=(x4−2x3+x2)+(x2−4x+4)+1=(x4−2x3+x2)+(x2−4x+4)+1
=(x2−x)2+(x−2)2+1=(x2−x)2+(x−2)2+1
Vì: {(x2−x)2≥0(x−2)2≥0{(x2−x)2≥0(x−2)2≥0⇒(x2−x)2+(x−2)2≥0⇒(x2−x)2+(x−2)2≥0
⇒(x2−x)2+(x−2)2+1>0⇒(x2−x)2+(x−2)2+1>0
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