bài 1: tính hợp lí
-24 ( - 3 - 7 ) - 7 (-24 -3 )
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1: 0,35*12,4=0,35*2,4+0,35*10=3,5+0,84=4,34
2: =0,1-2,34=-2,24
3: =5-2,9=2,1
4: \(=2,5\left(10,124-0,124\right)=10\cdot2,5=25\)
5: =-3/7+1/13
=-39/91+7/91
=-32/91
6: =-1/3+1/3=0
1−2−3+4+5−6−7+8+...+21−22−23+24+25
= (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (21 - 22 - 23 + 24) + 25=(1−2−3+4)+(5−6−7+8)+...+(21−22−23+24)+25
= 0 + 0 + ... + 0 + 25=0+0+...+0+25
= 25
1) \(35.18-5.7.28\)
\(=35.18-35.28\)
\(=35.\left(18-28\right)\)
\(=35.\left(-10\right)\)
\(=-350\)
2) \(45-5.\left(12+9\right)\)
\(=45-60-45\)
\(=\left(45-45\right)-60\)
\(=0-60\)
\(=-60\)
3) \(24.\left(16-5\right)-16.\left(24-5\right)\)
\(=\left(384-120\right)-\left(384-120\right)\)
\(=264-264\)
\(=0\)
4) \(13.\left(23+22\right)-3.\left(17+28\right)\)
\(=13.45-3.45\)
\(=\left(13-3\right).45\)
\(=10.45\)
\(=450\)
tk ủng hộ nha!!!!!!!!1
1/ 35 . 18 - 5 . 7 . 28
= 35 . 18 - 35 . 28
= 35 . (18 - 28)
= 35 . (-10)
= -350
2/ 45 - 5 . (12 + 9)
= 5 . 9 - 5 . 12 + 5 . 9
= 5 . (9 - 12 + 9)
= 5 . 6
= 30
3/ 24 . (16 - 5) - 16 . (24 - 5)
= 24 . 16 - 24 . 5 - 16 . 24 - 16 . 5
= (24 . 16 - 16 . 24) - (24 . 5 - 16 . 5)
= 0 - [5 . (24 - 16)]
= 0 - [5 . 8]
= 0 - 40
= -40
4/ 13 . (23 + 22) - 3 . (17 + 28)
= 13 . 45 - 3 . 45
= 45 . (13 - 3)
= 45 . 10
= 450
a)=3/4+6/13+7/13+8/45+0,25
=(6/13+7/13)+(3/4+0,25)+8/45
=1+1+8/45
=98/45
a)=3/4+6/13+7/13+8/45+1/4
=(3/4+1/4)+(6/13+7/13)+8/45
=1+1+8/45=2+8/45=90/45+8/45=98/45
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: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
b: \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)
\(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1+1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)
\(=\dfrac{2}{\sqrt{6}\left(\sqrt{6}+2\right)}=\dfrac{2}{6+2\sqrt{6}}=\dfrac{1}{3+\sqrt{6}}=\dfrac{3-\sqrt{6}}{3}\)
-24 ( - 3 - 7 ) - 7 (-24 -3 ) = 429
=429 nha bn