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a/ 3^21 > 2^31
b/ 2017^10 + 2017^9 <2018^10
chọn mình nha . Mình cũng học lớp 6 đó (>-<)
Bài 1:
1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)
\(=-12-18\)
=-30
2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)
\(=36-2020+2019-136-27\)
\(=1-100-27\)
\(=-126\)
3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)
\(=144-97-244+197\)
\(=-100+100=0\)
4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)
\(=-24\cdot13+24\cdot3\)
\(=24\cdot\left(-13+3\right)\)
\(=24\cdot\left(-10\right)=-240\)
5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)
\(=54+55+56+57+58-64-65-66-67-68\)
\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)
\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)
=-50
6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)
\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)
\(=-24\cdot5+16\cdot5\)
\(=5\cdot\left(-24+16\right)\)
\(=-5\cdot8=-40\)
7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)
\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)
\(=47\cdot50-23\cdot50\)
\(=50\cdot\left(47-23\right)\)
\(=50\cdot24=1200\)
8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)
\(=-31\cdot\left(47+52+1\right)\)
\(=-31\cdot100=-3100\)
Bài 2:
1) Ta có: \(-17-\left(2x-5\right)=-6\)
\(\Leftrightarrow-17-2x+5+6=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow-2x=6\)
hay x=-3
Vậy: x=-3
2) Ta có: \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow10-8+6x+4=0\)
\(\Leftrightarrow6x+6=0\)
\(\Leftrightarrow6x=-6\)
hay x=-1
Vậy: x=-1
3) Ta có: \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow-12-3x+21+18=0\)
\(\Leftrightarrow-3x+27=0\)
\(\Leftrightarrow-3x=-27\)
hay x=9
Vậy: x=9
4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)
\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)
\(\Leftrightarrow-2x-3=-3\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: x=0
5) Ta có: x(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3\right\}\)
6) Ta có: (x-2)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-1;3\right\}\)
Bài 1:
1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)
=−12−18=−12−18
=-30
2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27
=36−2020+2019−136−27=36−2020+2019−136−27
=1−100−27=1−100−27
=−126
Tớ chcs cậu học thật giỏi nha !
\(a,\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\frac{-3}{4}+\frac{-1}{4}+\frac{3}{7}+\frac{4}{7}+\frac{4}{9}\)
\(=-1+1+\frac{4}{9}\)
\(=\frac{4}{9}\)
\(b,\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(=\frac{-7}{9}\cdot1+\frac{52}{9}\)
\(=\frac{-7}{9}+\frac{52}{9}=\frac{45}{9}=5\)
Ý c) bn tự làm nha. Mk ko làm đc. Nhìn nó rắc rối lắm.
Chúc bn hc tốt nha !!! TỨ DIỆP THẢO
Em nến gõ bằng công thức toán học em nhé. Như vậy mọi người nới hiểu đúng đề bài để trợ giúp cho em một cách tốt nhất em ạ!
\(A=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+5^6\left(1+5+5^2\right)+...+5^{30}\left(1+5+5^2\right)=\)
\(=31\left(1+5^3+5^6+5^9+...+5^{30}\right)⋮31\)
\(\frac{2}{5}.\frac{19}{31}+\frac{2}{5}.\frac{43}{31}-2017\frac{4}{5}\)
\(=\frac{2}{5}.\left(\frac{19}{31}+\frac{43}{31}\right)-\frac{10089}{5}\)
\(=\frac{2}{5}.2-\frac{10089}{5}\)
\(=\frac{4}{5}-\frac{10089}{5}\)
\(=\frac{-10085}{5}\)
\(=-2017\)
CHÚC BẠN HỌC TỐT !!!!!!!!!!!!!!
\(\frac{2}{5}.\frac{19}{31}+\frac{2}{5}.\frac{43}{31}-2017\frac{4}{5}\)
\(=\frac{2}{5}\left(\frac{19}{31}+\frac{43}{31}\right)-\frac{10089}{5}\)
\(=\frac{2}{5}.2-\frac{10089}{5}\)
\(=\frac{4}{5}-\frac{10089}{5}=-2017\)