Rồi sao? đề bài?

\(4(x+1)^2-(2x-1)^2-8(x-1)(x+1)=11\)

\(\Leftrightarrow4\left(x^2+2x+1\right)-\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)

\(\Leftrightarrow4x^2+8x+4-4x^2+4x-1-8x^2+8=11\)

\(\Leftrightarrow-8x^2+12x+11=11\)

\(\Leftrightarrow-4x\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Ta có:

\(4\left(x+1\right)^2-\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\\ \Leftrightarrow4x^2+8x+4-4x^2+4x-1-8x^2+8=11\\ \Leftrightarrow-8x^2+12x=0\\ \Leftrightarrow-4x\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

( 2x - 1 )( x - 3 ) = 6

2x^2 - 6x - x + 3 = 6

2x^2 - 7x = 3 

x( 2x - 7 ) = 3 

Có 3 = 1 . 3 = 3 . 1 = -1 . -3 = -3 . -1

TH1 : x = 1 

=> 2x - 7 = 3

=> 2x = 10

=> x = 5

Nhưng ở đây x = 1 . Vậy loại trường hợp này 

TH2 : x = 3

=> 2x - 7 = 1

=> 2x = 8 

=> x = 4

Nhưng ở đây x = 3 . Vậy cũng loại trường hợp này 

TH3 : x = -1

Với x = -1 thì 2x - 7 = -3

=> 2x = 4 

=> x = 2 

Mà ở đây x = -1 . Vậy cũng loại trường hợp này

TH4 : x = -3

=> 2x - 7 = -1

=> 2x = 6

=> x = 3 

Vậy không tồn tại x 

HI,Name sophia

Hello name ?

The Mousea 

1+2+3+4+5+6+7+8+9

=(1+9)+(2+8)+(3+7)+(4+6)+5

=10+10+10+10+5

=10×4+5

=40+5

=45

Học tốt

= (1+9) + (2+8) + ( 3+7 ) + (4+6) + 5

= 10+10+10+10+5

= 10*4+5

=40+5

=45

=( 1+ 9) +( 2+8) +(3+ 7) +(4+ 6) + 5

=10 + 10 + 10 + 10 + 5

=10 x 4 + 5

=45

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)

\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

\(A=\frac{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}\)

\(A=\frac{1.101}{2.100}=\frac{101}{200}\)

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)

\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{3.4.5.6.....101}{2.3.4.5.....100}\)

\(A=\frac{1}{100}.\frac{101}{2}\)

\(A=\frac{101}{200}\)

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)

\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)

\(A=\frac{1.2...99}{2.3...100}.\frac{3.5...101}{2.3...100}\)

\(A=\frac{1}{100}.\frac{101}{2}\)

\(=\frac{101}{200}\)

\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)

\(B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}....\frac{1558}{1560}\)

\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(B=\frac{1.2.3...38}{2.3...39}.\frac{4.5...41}{3.4...40}\)

\(B=\frac{1}{39}.\frac{41}{3}\)

\(B=\frac{41}{117}\)

\(\frac{3}{5}+\frac{1}{6}+\frac{7}{30}=\frac{18}{30}+\frac{5}{30}+\frac{7}{30}=\frac{30}{30}=1\)

\(\frac{3}{4}-\frac{1}{2}+\frac{2}{5}=\frac{3}{4}-\frac{2}{4}+\frac{2}{5}=\frac{1}{4}+\frac{2}{5}=\frac{5}{20}+\frac{8}{20}=\frac{13}{20}\)

Chúc bạn học tốt!

\(\frac{3}{5}+\frac{1}{6}+\frac{7}{30}\)

\(=\frac{18}{30}+\frac{5}{30}+\frac{7}{30}\)

\(=\frac{18+5+7}{30}\)

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

\(=1\)

\(\frac{3}{4}-\frac{1}{2}+\frac{2}{5}\)

\(\frac{15}{20}-\frac{10}{20}+\frac{8}{20}\)

\(=\frac{15-10+8}{20}\)

\(=\frac{13}{20}\)

3/5 + 1/6 + 7/30 = 18/30 + 5/30 + 7/30 = 23/30 + 7/30 = 30/30 = 1

3/4 - 1/2 + 2/5 = 3/4 -2/4 + 2/5 = 1/4 + 2/5 = 5/20 + 8/20 = 13/20

ĐK: \(\left\{{}\begin{matrix}x\ne-y\\y\ge\dfrac{3}{2}\end{matrix}\right.\).

\(\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}=1\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-1=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-\dfrac{x+y}{x+y}=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y+3-x-y=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y+3=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\left(2y-3\right)=0\\2x-\sqrt{2y-3}=0\end{matrix}\right..\)

Đặt a = x, b = \(\sqrt{2y-3}\).

Hệ phương trình trở thành: \(\left\{{}\begin{matrix}a-b^2=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\2b^2-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\b\left(2b-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}a=0\\a=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y-3=\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y=\dfrac{13}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\y=\dfrac{13}{8}\end{matrix}\right.\end{matrix}\right..\)

Vậy hệ phương trình có nghiệm (x;y) \(\in\) \(\left\{\left(0;\dfrac{3}{2}\right),\left(\dfrac{1}{4};\dfrac{13}{8}\right)\right\}\).

A>B

mk nhắc rồi na

Anh qua câu hỏi của em đi, có ng trả lời mà, sao em hỏi nảy h anh ko trả lời

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}+\frac{1}{16}\)

 \(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)\)

\(+\left(\frac{1}{15}+\frac{1}{16}\right)\)

Vì \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}< 3\times\frac{1}{6}=\frac{1}{2}\)

\(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}< 3\times\frac{1}{9}=\frac{1}{3}\)

\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}< 3\times\frac{1}{12}=\frac{1}{4}\)

\(\frac{1}{15}+\frac{1}{16}< 3\times\frac{1}{15}=\frac{1}{5}\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3\times\frac{1}{2}=\frac{3}{2}\)

\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}+\frac{1}{16}< 1+\frac{3}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)

\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{15}+\frac{1}{16}< \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)

P.s Mình tịt rồi , bạn cố gắng giải ra nhá ^.^!!

\((6-4\frac{1}{2}):0,03+(3\frac{1}{20}-2,65):4+\frac{2}{5}\)

\((6-4,5):0,03+(3,05-2,65):4+0,4\)

\(=1,5:0,03+0,4:4+0,4\)

\(=50+(0,4\cdot2):4\)

\(=50+0,2\)

\(=50,2\)

Chúc bạn học tốt