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a) \(-2011-\left(200-2011\right)\)
\(=-2011-200+2011\)
\(=\left(-2011+2011\right)-200\)
\(=0-200\)
\(=-200\)
b) \(\left(-2\right)^2-\left(-2000\right)^0+\left(-1\right)^{2018}-\left|-20\right|\)
\(=4-1+1-20\)
\(=4-20\)
\(=-16\)
Bài 1 :
\(a)-2011-(200-2011)\)
\(=-2011-(200+2011)\)
\(=(-2011+2011)-200\)
\(=0-200=-200\)
\(b)(-2)^2-(-2000)^0+(-1)^{2018}-\left|-20\right|\)
\(=4-1+1-20\)
\(=4-20=-16\)
\(c)23\cdot18-23\cdot26+(-23)\cdot2\)
\(=23\cdot(18-26)+-(23\cdot2)\)
\(=23\cdot(-8)+(-46)\)
\(=-230\)
Bài 2 : Tìm số nguyên x biết :
\(a)3x-(-5)=20\)
\(\Rightarrow3x+5=20\)
\(\Rightarrow3x=20-5\)
\(\Rightarrow3x=15\Rightarrow x=5\)
\(b)3(x+2)=-4+(-2)^3\)
\(\Rightarrow3(x+2)=-4+(-8)\)
\(\Rightarrow3(x+2)=-12\)
\(\Rightarrow x+2=-12\div3\)
\(\Rightarrow x+2=-4\)
Tự tìm x câu b, và câu c,
Bài 3 tự làm
a) Ta có |x-5|=13
suy ra x-5=13 hoặc x-5=-13
suy ra x=18 hoặc x=-13
b) |x-3|<4
suy ra x-3<4 hoặc 3-x<4
suy ra x<7 hoặc x >-1
c) d) e) f) y) tương tự nha bạn!
2/3*4=5/4*y
=)5/4*y=2/3*4
=)5/4*y=8/3
=)y=(8/3)/(5/4)
=) y = (8/3)*(4/5)
=) y = 32/15
mà x + y =23
=)x =32-y
=)x= 32-32/15
=)x =448/15
=>(4-x:2)^3=1+2*(8-5)+1
=>(4-x:2)^3=2+2*3=8
=>4-x:2=2
=>x:2=2
=>x=4
\(\dfrac{4}{3}\) = \(\dfrac{20}{4}\)\(x\) - 1
5\(x\) - 1 = \(\dfrac{4}{3}\)
5\(x\) = \(\dfrac{4}{3}\) + 1
5\(x\) = \(\dfrac{7}{3}\)
\(x\) = \(\dfrac{7}{3}\) : 5
\(x\) = \(\dfrac{7}{15}\)
b,
2\(\times\)\(x\) = 3\(\times\) y
\(x\) = \(\dfrac{3}{2}\) \(\times\) y
y - \(\dfrac{3}{2}\) \(\times\) y = 5
\(y\) \(\times\) ( 1 - \(\dfrac{3}{2}\)) = 5
\(y\) \(\times\) - \(\dfrac{1}{2}\) = 5
\(y\) = 5 : (-\(\dfrac{1}{2}\))
\(y\) = - 10
\(x\) = y - 5 = -10 - 5 =-15
\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)
\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)
Bài 5 :
a) \(\dfrac{y}{4}=\dfrac{9}{y}\)
\(\Rightarrow y^2=36\left(y\ne0\right)\)
\(\Rightarrow y=\pm6\)
b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)
\(\Rightarrow\left(y+7\right)^2=100=10^2\)
\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)
c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)
\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)
\(\Rightarrow20-25y=3y+6\)
\(\Rightarrow28y=14\)
\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)
Bài 4 :
\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)
\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)
\(\frac{4}{5}+3y-\frac{1}{4}=\frac{23}{20}\)
\(=>\frac{4}{5}-\frac{1}{4}+3y=\frac{23}{20}\)
\(=>\frac{9}{20}+3y=\frac{23}{20}\)
\(=>3y=\frac{14}{20}\)
\(=>y=\frac{14}{60}=\frac{7}{30}\)
\(\frac{4}{5}+y.3-\frac{1}{4}=\frac{23}{20}\)
\(\frac{4}{5}+y.3=\frac{23}{20}+\frac{1}{4}\)
\(\frac{4}{5}+y.3=\frac{23}{20}+\frac{5}{20}\)
\(\frac{4}{5}+y.3=\frac{7}{5}\)
\(y.3=\frac{7}{5}-\frac{4}{5}\)
\(y.3=\frac{3}{5}\)
\(y=\frac{3}{5}:3\)
\(y=\frac{3}{5}.\frac{1}{3}\)
\(y=\frac{1}{5}\)