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Ta có: x2 – x – 12 = x2 – x – 16 + 4
= (x2 – 16) – (x – 4)
= (x – 4).(x + 4) – (x – 4)
= (x – 4).(x + 4 – 1)
= (x – 4).(x + 3)
Dạng 1:
a) \(x^4+y^2-2x^2y=\left(x^2-y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a-b\right)\left(a+b\right)\)
c) \(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)^2\)
d) \(a^3+b^3+c^3-3abc\)
\(=a^3+3a^2b+3ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ca-bc-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Dạng 2:
a) \(\left(7n-2\right)^2-\left(2n-7\right)^2\)
\(=\left(7n-2-2n+7\right)\left(7n-2+2n-7\right)\)
\(=\left(5n+5\right)\left(9n-9\right)\)
\(=45\cdot\left(n+1\right)\cdot\left(n-1\right)⋮3;5;9\) chứ không chia hết cho 7
Bạn xem lại đề.
b) \(n^3-n=n\left(n^2-1\right)=n\left(n-1\right)\left(n+1\right)\)
Vì \(n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên tích đó chia hết cho 2 và 3.
Mặt khác \(\left(2;3\right)=1\)
Do đó \(n\left(n-1\right)\left(n+1\right)⋮2.3=6\) ( đpcm
a) A = (-1.2axy^2)^3
A = (-1/2)^3.a^3x^3y^6
A = -1/8.a^3.x^3.y^6
- Hệ số: -1/8.a^3
các bn còn lại bn lm tương tự nha!
- Bậc: 9
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
Bài 1:
\(\frac{A}{x-1}+\frac{B}{x-2}=\frac{A\left(x-2\right)+B\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)
\(=\frac{Ax-2A+Bx-B}{x^2-3x+2}=\frac{\left(A+B\right)x-\left(2A+B\right)}{x^2-3x+2}\)
so sách với tử số vừa tìm dc với đề bài:
=> A+B=1
2A+B=-2
=>(2A+B)-(A+B)=-2-1
A=-3
=> B=1+3=4
b) sửa đề \(\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}=\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}\)
=> \(\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\frac{Ax^2+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(A+B\right)x^2+\left(C-B\right)x+\left(A-C\right)}{\left(x-1\right)\left(x^2+1\right)}\)
so sánh với tử số bên cạnh là \(x^2+2x-1\)
=>\(A+B=1\)
\(C-B=2\)
\(A-C=-1\)
=> \(A=1,B=0,C=2\)
bài 2:
quy đồng hai hạng tử đầu tiên:
=> \(\frac{x}{1-x^2}+\frac{y}{1-y^2}=\frac{x\left(1-y^2\right)+y\left(1-x^2\right)}{\left(1-x^2\right)\left(1-y^2\right)}=\frac{\left(x+y\right)\left(1-xy\right)}{\left(1-x^2\right)\left(1-y^2\right)}\)
từ xy+yz+xz=1=> 1-xy=z(x+y) thay vào biểu thức vừa tìm dc ta có:
\(\frac{\left(x+y\right)z\left(x+y\right)}{\left(1-x^2\right)\left(1-y^2\right)}=\frac{z\left(x+y\right)^2}{\left(1-x^2\right)\left(1-y^2\right)}\)
\(VT=\frac{z\left(x+y\right)^2}{\left(1-x^2\right)\left(1-y^2\right)}+\frac{z}{1-z^2}=z\left\lbrace\frac{\left(x+y\right)^2\left(1-z^2\right)+\left(1-x^2\right)\left(1-y^2\right)}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\right)\)
ta có:
\(\left(x+y\right)^2-z^2\left(x+y\right)^2+1-x^2-y^2+x^2y^2\)
=\(\left(x^2+2xy+y^2\right)-z^2\left(x+y\right)^2+1-x^2-y^2+x^2y^2\)
=\(\left(1+xy\right)^2-z^2\left(x+y\right)^2=\left(1+xy-xz-yz\right)\left(1+xy+xz+yz\right)\)
=\(4xy\)
thay vào biểu thức ban đầu:
\(z\cdot\frac{4xy}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\left(đpcm\right)\)
bài 3:
xếp hạng tổng k của dãy số:
\(a_{k}=\frac{k}{k^4+k+1}\)
=> \(a_{k}=\frac12\left\lbrace\frac{\left(k^2+k+1\right)-\left(k^2-k+1\right)}{\left(k^2-k+1\right)\left(k^2+k+1\right)}\right\rbrace=\frac12\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)\)
thay k=1,2,3,4,...,n)
=> \(S=\frac12\left\lbrace\left(\frac11-\frac13\right)+\left(\frac13-\frac17\right)+\cdots+\left(\frac{1}{n^2-n+1}-\right.\frac{1}{n^2+n+1}\right)\) S=\(\frac12\left(1-\frac{1}{n^2+n+1}\right)\)
\(S=\frac{n\left(n+1\right)}{2\left(n^2+n+1\right)}\)
B1:a)(3x-5)2-(3x+1)2=8
[(3x-5)+(3x+1)].[(3x-5)-(3x+1)]=8
(3x-5+3x+1)(3x-5-3x-1)=8
9x2-15x-9x2-3x-15x+25+15x+5+9x2-15x-9x2-3x+3x-5-3x-1=8
-36x+24=8
-36x=8-24=16
x=16:(-36)=\(\dfrac{-4}{9}\)
Bài 5:
a: \(=\left(xy-u^2v^3\right)\left(xy+u^2v^3\right)\)
b: \(=\left(2xy^2-3xy^2+1\right)\left(2xy^2+3xy^2-1\right)\)
\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)
Bài 6:
a: \(\left(a+b+c-d\right)\left(a+b-c+d\right)\)
\(=\left(a+b\right)^2+\left(c-d\right)^2\)
\(=a^2+2ab+b^2+c^2-2cd+d^2\)
b: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=a^2-2ad+d^2-b^2+2bc-c^2\)
1) A=\(\left(\frac34\cdot\frac45\cdot\frac56\right)\cdot\left(x^{n-1}\cdot x^{2n+1}\cdot x\right)\left(y^{2n+1}\cdot y^{n+1}\right)\)
\(A=\frac12\cdot x^{n-1+2n+1+1}\cdot y^{2n+1+n+1}\)
\(A=\frac12x^{3n+1}y^{3n+2}\)
hệ số : \(\frac12\)
bậc của đơn thức: (3n+1)+(3n+2)=6n+3
2) \(B=\left(\frac64\cdot\frac42\cdot\frac26\right)\cdot\left(x^{3-n}\cdot x^{4-n}\right)\left(y^{5-n}\cdot y^{6-n}\right)\)
\(B=1\cdot x^{3-n+4-n}\cdot y^{5-n+6-n}\)
\(B=x^{7-2n}\cdot y^{11-2n}\)
hệ số: 1
bậc của đơn thức: (7-2n)+(11-2n)=18-4n
3) \(C=\left(\frac{-4}{3}\cdot\frac67\cdot\frac{-1}{2}\right)\cdot\left(x^{2-n}\cdot x^{2n-3}\cdot x\right)\left(y\cdot y^{n-1}\cdot y\right)\)
\(C=\frac47\cdot x^{2-n+2n-3+1}\cdot y^{1+n-1+1}\)
\(C=\frac47x^{n}\cdot y^{n+1}\)
hệ số: \(\frac47\)
bậc của đơn thức: n+(n+1)=2n+1
4) \(D=\left(\frac15\cdot\frac43\cdot\frac{15}{7}\right)\cdot\left(x\cdot x^{n+1}\cdot x^{n}\right)\left(y^{n+1}\cdot y\cdot y^{n}\right)\)
\(C=\frac47\cdot x^{1+n+1+n}\cdot y^{n+1+1+n}\)
\(C=\frac47x^{2n+2}\cdot y^{2n+2}\)
hệ số: \(\frac47\)
bậc của đơn thức: (2n+2)+(2n+2)=4n+4
- Thu gọn:
- Hệ số: \(\frac{1}{2}\)
- Bậc: \((3n+1) + (3n+2) = 6n+3\)
2) Đơn thức \(B\)\(B = \frac{6}{4} x^{3-n} \cdot \frac{4}{2} x^{4-n} y^{5-n} \cdot \frac{2}{6} y^{6-n}\)\(A = \left( \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \right) \cdot \left( x^{n-1} \cdot x^{2n+1} \cdot x^1 \right) \cdot \left( y^{2n+1} \cdot y^{n+1} \right)\)
\(A = \frac{1}{2} x^{(n-1) + (2n+1) + 1} y^{(2n+1) + (n+1)}\)
\(A = \frac{1}{2} x^{3n+1} y^{3n+2}\)
- Thu gọn:
- Hệ số: \(1\)
- Bậc: \((7-2n) + (11-2n) = 18-4n\)
3) Đơn thức \(C\)\(C = \frac{-4}{3} x^{2-n} y \cdot \frac{6}{7} x^{2n-3} y^{n-1} \cdot \frac{-1}{2} xy\)\(B = \left( \frac{6}{4} \cdot \frac{4}{2} \cdot \frac{2}{6} \right) \cdot \left( x^{3-n} \cdot x^{4-n} \right) \cdot \left( y^{5-n} \cdot y^{6-n} \right)\)
\(B = 1 \cdot x^{(3-n) + (4-n)} y^{(5-n) + (6-n)}\)
\(B = x^{7-2n} y^{11-2n}\)
- Thu gọn:
- Hệ số: \(\frac{4}{7}\)
- Bậc: \(n + (n+1) = 2n+1\)
4) Đơn thức \(D\)\(D = \frac{1}{5} x y^{n+1} \cdot \frac{4}{3} x^{n+1} y \cdot \frac{15}{7} x^n y^n\)\(C = \left( \frac{-4}{3} \cdot \frac{6}{7} \cdot \frac{-1}{2} \right) \cdot \left( x^{2-n} \cdot x^{2n-3} \cdot x^1 \right) \cdot \left( y^1 \cdot y^{n-1} \cdot y^1 \right)\)
\(C = \frac{4}{7} x^{(2-n) + (2n-3) + 1} y^{1 + (n-1) + 1}\)
\(C = \frac{4}{7} x^n y^{n+1}\)
\(D = \left( \frac{1}{5} \cdot \frac{4}{3} \cdot \frac{15}{7} \right) \cdot \left( x^1 \cdot x^{n+1} \cdot x^n \right) \cdot \left( y^{n+1} \cdot y^1 \cdot y^n \right)\)
\(D = \frac{4}{7} x^{1 + (n+1) + n} y^{(n+1) + 1 + n}\)
\(D = \frac{4}{7} x^{2n+2} y^{2n+2}\)
- Thu gọn:
- Hệ số: \(\frac{1}{2}\)
- Bậc: \((3n+1) + (3n+2) = \mathbf{6n+3}\)
2) Đơn thức \(B\)\(B = \frac{6}{4}x^{3-n} \cdot \frac{4}{2}x^{4-n}y^{5-n} \cdot \frac{2}{6}y^{6-n}\)\(A = \left( \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \right) \cdot (x^{n-1} \cdot x^{2n+1} \cdot x) \cdot (y^{2n+1} \cdot y^{n+1})\)
\(A = \frac{1}{2}x^{(n-1) + (2n+1) + 1} \cdot y^{(2n+1) + (n+1)}\)
\(A = \frac{1}{2}x^{3n+1}y^{3n+2}\)
- Thu gọn:
- Hệ số: \(1\)
- Bậc: \((7-2n) + (11-2n) = \mathbf{18-4n}\)
3) Đơn thức \(C\)\(C = \frac{-4}{3}x^{2-n}y \cdot \frac{6}{7}x^{2n-3}y^{n-1} \cdot \frac{-1}{2}xy\)\(B = \left( \frac{6}{4} \cdot \frac{4}{2} \cdot \frac{2}{6} \right) \cdot (x^{3-n} \cdot x^{4-n}) \cdot (y^{5-n} \cdot y^{6-n})\)
\(B = 1 \cdot x^{(3-n) + (4-n)} \cdot y^{(5-n) + (6-n)}\)
\(B = x^{7-2n}y^{11-2n}\)
- Thu gọn:
- Hệ số: \(\frac{4}{7}\)
- Bậc: \(n + (n+1) = \mathbf{2n+1}\)
4) Đơn thức \(D\)\(D = \frac{1}{5}xy^{n+1} \cdot \frac{4}{3}x^{n+1}y \cdot \frac{15}{7}x^n y^n\)\(C = \left( \frac{-4}{3} \cdot \frac{6}{7} \cdot \frac{-1}{2} \right) \cdot (x^{2-n} \cdot x^{2n-3} \cdot x) \cdot (y \cdot y^{n-1} \cdot y)\)
\(C = \frac{4}{7}x^{(2-n) + (2n-3) + 1} \cdot y^{1 + (n-1) + 1}\)
\(C = \frac{4}{7}x^n y^{n+1}\)
\(D = \left( \frac{1}{5} \cdot \frac{4}{3} \cdot \frac{15}{7} \right) \cdot (x \cdot x^{n+1} \cdot x^n) \cdot (y^{n+1} \cdot y \cdot y^n)\)
\(D = \frac{4}{7}x^{1 + (n+1) + n} \cdot y^{(n+1) + 1 + n}\)
\(D = \frac{4}{7}x^{2n+2}y^{2n+2}\)
A = (3/4).(4/5).(5/6)x^(n-1+2n+1+1)y^(2n+1+n+1)
= 1/2 x^(3n+1)y^(3n+2)
Hệ số là 1/2
Bậc là (3n + 1) + (3n + 2) = 6n + 3
B = (6/4).(4/2).(2/6)x^(3-n+4-n)y^(5-n+6-n)
= x^(7-2n)y^(11-2n)
Hệ số là 1
Bậc là (7 - 2n) + (11 - 2n) = 18 - 4n
C = (-4/3).(6/7).(-1/2)x^(2-n+2n-3+1)y^(1+n-1+1)
= 4/7 x^n y^(n+1)
Hệ số là 4/7
Bậc là n + (n + 1) = 2n + 1
D = (1/5).(4/3).(15/7)x^(1+n+1+n)y^(n+1+1+n)
= 4/7 x^(2n+2)y^(2n+2)
Hệ số là 4/7
Bậc là (2n + 2) + (2n + 2) = 4n + 4