Ta tách ra làm 2 ý nhé:

* \(1,11+0,19-1,32-\left(\frac{1}{2}+\frac{1}{3}\right):2\)

\(=1,3-1,32-\frac{5}{6}=-0,02-\frac{5}{6}=-\frac{1}{50}-\frac{5}{6}=-\frac{64}{75}\)

* \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}\)

\(\left(3\frac{5}{8}-\frac{1}{2}\right):2\frac{23}{26}=3\frac{1}{8}:2\frac{23}{26}=1\frac{1}{12}=\frac{13}{12}\)

VẬY TA CÓ: \(\frac{-64}{75}< x< \frac{13}{12}\)mà  \(\left(x\inℕ\right)\)

\(\Rightarrow\)ta có x là số nguyên nằm giữa -0.8533... và 1,0833...

vậy ta có x là các số nguyên 0 và 1 

MK KO CHẮC CHO LẮM NÊN BN CÓ THỂ THAM KHẢO Ý KIẾN MẤY BN KHÁC NHÉ.

K MK NHA. CHÚC BN HỌC TỐT. ^_^

fqaeggg

a) \(\frac{1,11+0,19-12,2}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)

Đề đúng:

\(\frac{1,11+0,19-1,3.2^6}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)

\(\frac{1,3-1,3.64}{0,39}-x=\left(\frac{3}{6}+\frac{2}{6}\right).\frac{1}{2}\)

\(\frac{1,3.\left(-63\right)}{0,39}-x=\frac{5}{6}.\frac{1}{2}\)

\(\frac{-81,9}{0,39}-x=\frac{5}{12}\)

\(\left(-210\right)-x=\frac{5}{12}\)

\(x=\left(-210\right)-\frac{5}{12}\)

\(x=\frac{-2520}{12}-\frac{5}{12}\)

\(x=\frac{-2525}{12}\)

b) \(x:\left(3\frac{1}{2}-5\frac{1}{6}\right)=4\frac{1}{5}-6\frac{2}{3}\)

\(x:\left(\frac{7}{2}-\frac{31}{6}\right)=\frac{21}{5}-\frac{20}{3}\)

\(x:\left(\frac{21}{6}-\frac{31}{6}\right)=\frac{63}{15}-\frac{100}{15}\)

\(x:\frac{-10}{6}=\frac{-37}{15}\)

\(x.\frac{6}{-10}=\frac{-37}{5}\)

\(x.\frac{-6}{10}=\frac{-37}{5}\)

\(x.\frac{-3}{5}=\frac{-37}{5}\)

\(x=\frac{-37}{5}:\frac{-3}{5}\)

\(x=\frac{-37}{5}.\frac{5}{-3}\)

\(x=\frac{-37}{-3}\)

\(x=\frac{37}{3}\)

Sai đề bài

Lẽ ra câu a chỉ chia 2 thì chia 22

Sai đề bài mina-san 

thank you vì đã trả lời hihi

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

\(\left(x+2\right)^2=\left(2x-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-1\\x+2=-\left(2x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-2\\x+2=-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\x+2x=1-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)

(x+2)2=(2x−1)2⇔(x+2)2−(2x−1)2=0⇔[x+2−(2x−1)][x+2+2x−1]=0⇔(x+2−2x+1)(x+2+2x−1)=0⇔(−x+3)(3x+1)=0⇔[−x+3=03x+1=0​⇔[−x=−33x=−1​
 

=) x = 3 hoặc -1/3.

Chúc bạn hok tốt

9.6 x 45 .6 + 3.2 x 53.1 x 3 + 4.8 x 1.3 x 2 = 960

mình biết làm 

cách giải bạn ơi !

B=1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 +5 x 6 + ........... + 2015 x 2016

3xB=1 x 2x3 + 2 x 3x3 + 3 x 4x3 + 4 x 5x3 +5 x 6x3 + ........... + 2015 x 2016x3

=1x2x3+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+2014x2015(2016-2013)+2015x2016(2017-2014)

=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+..+2014x2015x2016-2013x2014x2015+2015x2016x2017-2014x2015x2016

các số hạng tự triệt tiêu nhau

còn lại (2015x2016x2017)

B= (2015x2016x2017)/3=2015x672x2017

2015=5.43

672=.3.4.7.8

=> B chia hết cho 2,4,6,7,8 

không chia hết cho 9

a)
=\(2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1\)
=\(2x=1+\dfrac{1}{2}=1.5\)
=\(x=1.5:2=0.75\)
b)
=\(x^2=0+2=2\)
TH1:\(x=2\)
TH2:\(x=-2\)

Bài 1:

a: \(2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}\)

=>\(2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}\)

=>\(2x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)

=>x=2/2=1

b: \(x^2-2=0\)

=>\(x^2=2\)

=>\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

Bài 2:

a: \(A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)

\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)

\(=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):\left(2\dfrac{23}{26}\right)\)

\(=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)

\(=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}\)

b: A<x<B

=>\(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

mà \(x\in Z\)

nên \(x\in\left\{-9;-8;...;0;1\right\}\)