Bài 1: Tìm
BC(24;8;20) và nhỏ hơn 500
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\(=\dfrac{-23-28}{24}=\dfrac{-51}{24}=\dfrac{-17}{8}\)
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: \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)
\(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1+1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)
\(=\dfrac{2}{\sqrt{6}\left(\sqrt{6}+2\right)}=\dfrac{2}{6+2\sqrt{6}}=\dfrac{1}{3+\sqrt{6}}=\dfrac{3-\sqrt{6}}{3}\)
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)
= (125 + 875)*3
= 1000*3
= 3000
b)
= (38 +62)*24
= 100*4
= 400
~~Học tốt nha~~
a. phai la 125x3+875x3 moi dung nhe!
= 3( 125+875)
= 3x 1000
= 3000
b. 38x24+62x24
=24(38+62)
= 24x 100
= 2400
Chuc e hoc gioi nhe!!!
Bài 2: Tìm x:
a. (x + 1) + (x + 2) + (x + 3) = 24 (x + x + x) + (1 +2 + 3) = 24 x × 3 + 6 = 24 x × 3 = 24 - 6 x × 3 = 18 x = 18 : 3 x = 6 | b. x + x + 8 = 24 2 × x + 8 = 24 2 × x = 24 - 8 2 × x = 16 x = 16 : 2 x = 8 |
24.(56-45)+56.(-24-45)
= 24.56-24.45-56.24-56.45
= (24.56-56.24)-(24.45+56.45)
= 0-45.(24+56)
= -45.80
= -3600
24 . (56 - 45) + 56(-24 - 45)
= 24 . 56 - 24 . 45 + 56 . (-24) - 56 . 45
= [24 . 56 + 56 . (-24)] - (24 . 45 + 56 . 45)
= 56 . [24 + (-24)] - 45(24 + 56)
= 56 . 0 - 45 . 80 = 0 - 3600 = -3600
m-24-x+24+x với x=37 m=72
72-24-37+24+37
=48-37+24+37
=11+24+37
=35+37
=72 nha bn
thay x=37 và m=72 vào biểu thức:
72-24-37+24+37=72
Ai k mk mk sẻ k lại
x=240;480;120;360
Lời giải:
$24=2^3.3$
$8=2^3$
$20=2^2.5$
$\Rightarrow BCNN(24,8,20)=2^3.3.5=120$
$\Rightarrow BC(24,8,20)\in \left\{0; 120; 240; 360; 480; 600;...\right\}$
Bội chung của $24,8,20$ nhỏ hơn $500$ là:
$\left\{0; 120; 240; 360; 480\right\}$