Tìm x: x0x04 + 40x0x + x040x=10101 x 11 x 2

Đáp án bằng X= 9 

biết rồi hỏi làm gì nữa

x0x04+40x0x+x040x=yyyyyy

Thay số:40404+40404+40404=121212

vậy yyyyyy=121212

ko hiểu câu hỏi

tìm x nha mg thông cảm nha

hihi tui quên viết đầu bài nha

tìm x nha

chịu

 ko trả lời được

E=\(\frac{5}{1.2}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

E.\(\frac{1}{7}\)=\(\frac{5}{1.2.7}+\frac{4}{1.11.7}+\frac{3}{11.2.7}+\frac{1}{2.15.7}+\frac{13}{15.4.7}\)

E.\(\frac{1}{7}\)=\(\frac{5}{7.2}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

E.\(\frac{1}{7}\)=\(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

E.\(\frac{1}{7}\)=\(\frac{1}{2}-\frac{1}{28}\)

E.\(\frac{1}{7}=\frac{13}{28}\)

E=\(\frac{13}{28}:\frac{1}{7}=\frac{13}{4}\)

\(E=\frac{5}{1.2}+\frac{1}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(E=\frac{5}{2}+\frac{1}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\)

\(E=\frac{5}{2}+\left(\frac{1}{11}+\frac{3}{22}\right)+\left(\frac{1}{30}+\frac{13}{60}\right)\)

\(E=\frac{5}{2}+\left(\frac{2}{22}+\frac{3}{22}\right)+\left(\frac{2}{60}+\frac{13}{60}\right)\)

\(E=\frac{5}{2}+\frac{5}{22}+\frac{15}{60}\)

\(E=\frac{55}{22}+\frac{5}{22}+\frac{1}{4}\)

\(E=\frac{60}{22}+\frac{1}{4}\)

\(E=\frac{30}{11}+\frac{1}{4}\)

\(E=\frac{120}{44}+\frac{11}{44}\)\(=\frac{131}{44}\)

k mình nha chúc bạn học giỏi

    \(\frac{1}{2.15}+\frac{3}{2.11}+\frac{4}{1.11}+\frac{5}{1.2}\)

\(=\frac{1}{30}+\left(\frac{3}{22}+\frac{4}{11}\right)+\frac{5}{2}\)

\(=\frac{1}{30}+\frac{1}{2}+\frac{5}{2}\)

\(=\frac{1}{30}+3\)

\(=\frac{91}{30}\)

= 3 \(\frac{1}{30}\)Hoặc  =\(\frac{91}{30}\)

ta có 

1/15.2=1/30=11/330

3/2.11=3/22=45/330

4/1.11=4/11=120/330

5/1.2=5/2=825/330

=>11/330+45/330+120/330+825/330

=11+45+120+825/330

=1001/330=91/330

Đặt \(A=\frac{1}{2\cdot15}+\frac{3}{2\cdot11}+\frac{4}{1\cdot11}+\frac{5}{1\cdot2}\)

\(\Leftrightarrow\frac{A}{7}=\frac{1}{14\cdot15}+\frac{3}{11\cdot14}+\frac{4}{7\cdot11}+\frac{5}{7\cdot2}\)

\(\Leftrightarrow\frac{A}{7}=\frac{1}{14}-\frac{1}{15}+\frac{1}{11}-\frac{1}{14}+\frac{1}{7}-\frac{1}{11}+\frac{1}{2}-\frac{1}{7}\)

\(\Leftrightarrow\frac{A}{7}=-\frac{1}{15}+\frac{1}{2}=\frac{1}{2}-\frac{1}{15}=\frac{15-2}{30}=\frac{13}{70}\)

hay \(A=\frac{13\cdot7}{70}=\frac{91}{70}=\frac{13}{10}\)

 bÀI LÀM

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)