Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: \(A=y^3-4y^2+2y-8-y^3+4y^2-\dfrac{1}{2}y+2\)
\(=\dfrac{3}{2}y-6=\dfrac{3}{2}\cdot\dfrac{-2}{3}-6=-1-6=-7\)
b: \(x+1=5\)
\(B=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2\)
=x-2
=4-2
=2
c: \(C=\dfrac{1}{229}\cdot\left(6+\dfrac{3}{433}\right)-\dfrac{1}{299}\cdot\dfrac{432}{433}-\dfrac{1}{229}\cdot\dfrac{4}{433}\)
\(=\dfrac{1}{299}\left(6+\dfrac{3}{433}-\dfrac{432}{433}-\dfrac{4}{333}\right)\)
\(=\dfrac{-5}{299}\)
bai1
\(3a\left(2+b\right)-a\left(1-b\right)-4ab=6a+3ab-a+ab-4ab=5a=\frac{5}{229}\)
bai3
\(M=4\left(X-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)=\)
\(4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x=-24\)
bai 4
\(\text{a(x-y)+b(y-x)}=\left(x-y\right)\left(a-b\right)\)
bai 5
ta co cong thuc tinh tong 1+2+3+4+5+...+150=\(\frac{\left(1+150\right)150}{2}=11325\)
a11325
bai 6
\(p=x\left(5x+15y\right)-5y\left(3x-2y\right)-5y^2+10\)
\(=5x^2+15xy-15xy+10y^2-5y^2+10=5x^2+5y^2+10=5\left(x^2+y^2\right)+10\)
ta nhan thay rang de P=10 thi (x2+y2)=0 suy ra x=y=0
P=0 thi (x2+y2)= -2 ma so chinh phuong bao gioi cung lon hon 0 nen truong hop nay vo nghiem de thoa man
3/229 x ( 2 + 1/433 ) - 1/229 x 432/433 - 4/229x433
= ( 3/229 x 867/433 ) - ( 432/99157 - 4/99157 )
= 2601/99157 - 432/99157 - 4/99157
= 5/229
\(M=\frac{3}{299}\cdot\left(2+\frac{1}{433}\right)-\frac{1}{299}\cdot\frac{432}{433}-\frac{4}{229.433}\)
\(M=\left(\frac{3}{229}\cdot\frac{867}{433}\right)-\left(\frac{432}{99157}-\frac{4}{99157}\right)\)
\(M=\frac{2601}{99157}-\frac{432}{99157}-\frac{4}{99157}\)
\(M=\frac{5}{229}\)