Bài 2:Tìm x bt
a)1200:24-(17-x)=36
b)x+\(\dfrac{1}{2}.x-25\%.x=10\)
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Bài 1
a) \(5\times72\times10\times2=\left(5\times2\times10\right)\times72=100\times72=7200\)
b) \(40\times125=5\times\left(8\times125\right)=5\times1000=5000\)
c) \(16\times6\times25=4\times4\times6\times25=\left(4\times6\right)\times\left(4\times25\right)=24\times100=2400\) Bài 2:
a) \(24\times57+43\times24=24\times\left(57+43\right)=24\times100=2400\)
b) \(12\times19+80\times12+12=12\times\left(19+80+1\right)=12\times100=1200\)
c) \(\left(36\times15\times169\right)\div\left(5\times18\times13\right)\)
\(=\left(18\times2\times3\times5\times13\times13\right)\div\left(5\times18\times13\right)\)
\(=\left(2\times3\times13\right)\times\left(18\times5\times13\right)\div\left(5\times18\times13\right)\)
\(=2\times3\times13\)
\(=78\)
d) \(\left(44\times52\times60\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times11\times4\times13\times4\times15\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times4\times4\right)\times\left(11\times13\times15\right)\div\left(11\times13\times15\right)\)
\(=4\times4\times4\)
\(=64\)
Bài 3:
a) \(x-280\div35=5\times54\)
\(x-8=270\)
\(x=270+8\)
\(x=278\)
b) \(\left(x-280\right)\div35=54\div4\)
\(\left(x-280\right)\div35=\dfrac{27}{2}\)
\(x-280=\dfrac{27}{2}\times35\)
\(x-280=\dfrac{945}{2}\)
\(x=\dfrac{945}{2}+280\)
\(x=\dfrac{1505}{2}\)
c) \(\left(x-128+20\right)\div192=0\)
\(x-128+20=0\times192\)
\(x-128+20=0\)
\(x-128=0-20\)
\(x-128=-20\)
\(x=-20+128\)
\(x=108\)
d) \(4\times\left(x+200\right)=460+85\times4\)
\(4\times\left(x+200\right)=460+340\)
\(4\times\left(x+200\right)=800\)
\(x+200=800\div4\)
\(x+200=200\)
\(x=200-200\)
\(x=0\)
Bài 4:
a) \(\dfrac{7}{12}-\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
b) \(\dfrac{8}{11}+\dfrac{19}{11}=\dfrac{27}{11}\)
c) \(\dfrac{3}{8}+\dfrac{5}{12}=\dfrac{9}{24}+\dfrac{10}{24}=\dfrac{19}{24}\)
d) \(\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{9}{12}+\dfrac{7}{12}=\dfrac{16}{12}=\dfrac{4}{3}\)
Bài 5:
a) \(x-\dfrac{6}{7}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}+\dfrac{6}{7}\)
\(x=\dfrac{47}{14}\)
b) \(\dfrac{12}{7}\div x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\dfrac{12}{7}\div x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\dfrac{12}{7}\div x=\dfrac{11}{15}\)
\(x=\dfrac{12}{7}\div\dfrac{11}{15}\)
\(x=\dfrac{180}{77}\)
a. 1200: 24 - ( 17 - x) = 36
50 - ( 17- x) = 36
17 - x = 50 - 36
17 - x = 14
x = 17 - 14
x = 3
1. a) 2 + (-25) + 41 + (-2) + 25 + (-41)
= 2 - 25 + 41 - 2 + 25 -41
= ( 2 -2 ) + ( 25 - 25 ) + ( 41 - 41 )
= 0 + 0 + 0
= 0
b) 10 + (-17) + 5 + (-7) + 17 + (-15)
= 10 - 17 + 5 - 7 + 17 - 15
= ( 10 + 5 -15 ) + ( 17 -17 ) - 7
= 0 + 0 - 7
= -7
c) (-22) + (-14) + 17 + (-24) + 13 + 30
= -22 - 14 + 17 - 24 + 13 + 30
= ( -22 - 14 - 24 ) + ( 17 + 13 + 30)
= -60 + 60
= 0
2 a) -22 < x < 23
x thuộc { -21 ;-20;-19;-18;-17;-16;-15;-14;-13;-12;-11;-10;-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21;22}
Tổng các số nguyên trên là
( -21 ) + ( -20 ) + ( -19 ) + ..... + 22
= 22 + [( -21 ) +21 ] + [( -20) + 20] + ..... 0
= 22 + 0 + 0 + ... + 0
= 22
b) -36 < x < 34
Tương tụ câu a)
nguyễn hiền phần a trình bày sai r nhé lẽ ra là
a) 2 + (-25) + 41 + (-2) + 25 + (-41)
= [(2 + (-2)] + [(-25) + 25] + [41 + (-41)]
= 0 + 0 + 0
= 0
k dùm nhem các anh ; chị ; em
Bài 1:
a: BCNN(10;12)=60
b: BCNN(24;10)=120
c: BCNN(4;14;26)=364
d: BCNN(6;8;10)=120
Ta có : 1200:24-(17-x)=36
=> 50-(17-x)=36
=>17-x=50-36
=>17-x=14
=>x=17-14=3
50 - (17 - x ) = 36
17 - x = 50-36= 14
x = 17 - 14 = 3
Vậy x = 3
\(a,\) ta có :
\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)
\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)
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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :
\(B=\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\left(\sqrt{x}-1\right)+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\sqrt{x}+1+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)
vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)
để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
vậy để \(B=2\) thì \(x=4\)
a) \(1200:24-\left(17-x\right)=36\)
\(50-\left(17-4\right)=36\)
\(50-\left(17-x\right)=36\)
\(17-x=50-36\)
\(17-x=14\)
\(x=17-14\)
\(x=3\)
Vậy \(x=3\) là giá trị cần tìm
b) \(x+\dfrac{1}{2}.x-25\%.x=10\)
\(x+\dfrac{1}{2}x-\dfrac{25}{100}x=10\)
\(x+\dfrac{1}{2}x-\dfrac{1}{4}x=10\)
\(x\left(1+\dfrac{1}{2}-\dfrac{1}{4}\right)=10\)
\(x\left(1+\dfrac{1}{2}+\dfrac{-1}{4}\right)=10\)
\(x.\dfrac{5}{4}=10\)
\(x=10:\dfrac{5}{4}\)
\(x=8\)
Vậy \(x=8\) là giá trị cần tìm