bai 2: cho A=3 -2x²; B = 2x²-1. Tìm C biết C - A = B
bai 1 tinh
a)(x+3)^2
b)(2x+1)^2-(2x+3)(2x-3) giup em nhanh nhe
a) \(=x^2+6x+9\)
b) \(=4x^2+4x+1-4x^2+9=4x+10\)
a) \(\left(x+3\right)^2=x^2+6x+9\)
b) \(\left(2x+1\right)^2-\left(2x+3\right)\left(2x-3\right)=4x^2+4x+1-4x^2+9\)
\(=4x+10\)
bai 166 a) 6x^2 -11x +3 phan tich cac da thuc sau thanh nhan tu
b) 2x^+3x-27
c) 2x^2-5xy-3y^2
bai 167 a) x^3+2x-3 b) x^3-7x+6 c)x^3 +5x^2 +8x +4 d) x^3 -9x^2 +6x +16
e)x^3-x^2-x-2 g ) x^3+x^2-x+2 h)x^3 -6x^2-x+30
bai 169 a) 27x^3-27x^2 +18x-4
b)2x^3-x^2+5x+3
c)(x^2-3)^2+16
Dài 166
b) 2x2+3x-27=2x2-6x+9x-27=2x(x-3)+9(x-3)=(x-3)(2x+9)
bai 1:tim x
a)x+x2-x3-x4=0
b)2x3+3x2+2x+3=0
c)x4-2x3+10x2-20x=0
bai 2:tim cap so (x,y) biet
xy+3x-4y=12
bai 3:voi gia tri nao cua x
a)x3-x2+3x-3>0
b)x3+x3+9x+9<0
c)4x3-14x2+6x-21<0
d)x2(2x2+3)+2x2+3>0
bai 4:CMR hieu binh phuong 2 so le lien tiep thi chia het cho 8
bai 1:tim GTNN cua bieu thuc
A=x2+3x+7
B=(x-2)(x-5)(x2-7x-10)
bai 2:tim GTLN cua bieu thuc
A=11-10x-x2
B=[x-4](2-[x-4])
bai 3:tim x,y sao cho
A=2x2+9y2-6xy-6x-12y+2016 co GTNN
B=-x2+2xy-4y2+2x+10y-8 co GTLN
bai 4 :
a)cho x+y=3;x2+y2=5.tinh x3+y3
b)cho x-y=5;x2+y2=15.tinh x3-y3
giải giúp nhé
tìm x
bài 1
[x+2]+[3-4x]=16
bai 2
4[2x-3]+1=17
bai 3
[2x-1]-[3x+7]=0
bai 4
[5-3x]-2x = -1
bai 5
[4-5x]+3]-2=17
bai 6 [x+1]+[2x-3]-[1-3x]=2
nhớ đưa ra cách giải nhé (chú thích : dấu [ hoặc ] là giá trị tuyệt đối)
bai 1 tinh
a ) (x +3) (x^2 + 3x -5)
b ) (2x -5 )^2 - (2x +5 )^2
c) (x + 5) ^2 - (x +5 ) (x- 1)
bai 2 phan tich
a) 4x^2 - 6x
b) x ^4 + 2x^3 -4x -4
c) x^3 - y^3 -4x - 4y
d) (5x -4 )^2 -49 x^2
Bai 1: tinh
a(xy-2)(x^3-2x-6)
b(5x^3-x^2+2x-3)(4x^2-x+2)
\(\left(xy-2\right)\left(x^3-2x-6\right)=x^4y-2x^2y-6xy-2x^3+4x+12\)
\(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^5-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6\)
\(=20x^5-9x^4+19x^3-16x^2+7x-6\)
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
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
\(=3\left[\left(a-b\right)^2-4c^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
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 :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
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) \(3a^2-6ab+3b^2-12c^2\)
\(=\left(3a^2-6ab+3b^2\right)-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}\right)^2-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}+\sqrt{12c}\right)\left(\sqrt{3a}+\sqrt{3b}-\sqrt{12c}\right)\)
bai 1 tim x a) ( 4 1/2 - 2x) . 1 4/61 = 6 1/2 b) 2x + 1/2 = -5/3