Rút gọn biểu thức:
\(\left(3x-2y\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)+\left(y+2x\right)^3\)
rút gọn rồi tính giá trị biểu thức tại x=1; y=2
A= \(\dfrac{6x^3-4x^2y+2x^2}{x\left(3x+y\right)\left(3x-y\right)}\)
\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)
RÚT GỌN BIỂU THỨC
\(2x.\left(2x-1\right)-3x.\left(x+3\right).\left(x-3\right)-4x.\left(x+1\right)^2\)
cho hình thang giác vuông ABCD có
;góc A= D (=90) độ gọi M là trung điểm của bc
CMR: BAM=CDM
làm giúp mình ik mình lm cho
cho hình thang giác vuông ABCD có
;góc A= D (=90) độ gọi M là trung điểm của bc
CMR: BAM=CDM
lm giúp mình ikminhf lm cho
=4x2-2x-3x3-27x-4x3-8x2-4x
=-7x3-33x-4x2
=-x.(7x2+4x+33)
Rút gọn biểu thức:
\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)
Ta có: \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)
\(=x^3+125-x^3-9x^2-27x-27+x^3-8-x^3+3x^2-3x+1\)
\(=-6x^2-30x+91\)
B1: tính nhẩm
\(101^2\)
\(72^2+56.72+28^2\)
\(45^2\)
\(34.46\)
B2 : Rút gọn
\(A=\left(x+5y\right)^2-5\left(x-2y\right)^2\)
\(B=\left(3x-4\right)^2-2\left(x+11\right)\left(x-11\right)\)
\(C=\left(3x-2y\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)+\left(y+2x\right)^3\)
\(D=\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)
\(E=-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)
Tính
\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
Giups mk vs nha !! thanks
Rút gọn các biểu thức sau :
a) \(6x^n\left(x^2-1\right)+2x^3\left(3x^{n+1}+1\right)\)
b) \(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
c) \(x^{n-3}\left(x-y\right)+y\left(x^{n-3}+x^{n-3}y^{n-1}\right)\)
Rút gọn biểu thức :
\(\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
Lời giải :
\(\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{4}\cdot\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{4}\cdot\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{4}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{4}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{4}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{4}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\frac{3^{64}-1}{4}\)
Rút gọn và tính giá trị biểu thức :
\(N=\) \(\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)\(với\)\(a=\dfrac{1}{2};b=-3\)
\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)=\left(a-3b-a-3b\right)\left(a-3b+a+3b\right)-\left(ab-2a-b+2\right)=\left(-6b\right).2a-ab+2a+b-2=2a+b-13ab-2\)
Thay \(a=\dfrac{1}{2};b=-3\) vào N ta được: \(N=2a+b-13ab-2=2.\dfrac{1}{2}-3-13.\dfrac{1}{2}.\left(-3\right)-2=\dfrac{31}{2}\)
Ta có: \(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)-1-3-2\)
\(=\dfrac{27}{2}\)
\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
Thay \(a=\dfrac{1}{2};b=-3\) vào bt N được
\(N=\left(-13\right)\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{31}{2}\)
Vậy: Giá trị của N tại \(a=\dfrac{1}{2};b=-3\) là \(\dfrac{31}{2}\)
rút gọn biểu thức( không tính cái gạch ở cuối nha)\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)
\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4
Phân tích các đa thức sau thành nhân tử:
\(A=4x^2+6x\). \(B=\left(2x+3\right)^2-x\left(2x+3\right)\). \(C=\left(9x^2-1\right)-\left(3x-1\right)^2\).
\(D=x^3-16x\). \(E=4x^2-25y^2\). \(G=\left(2x+3\right)^2-\left(2x-3\right)^2\).
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)