Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: x+1/4=1/5
=>x=1/5-1/4
=>x=-1/20
b: x-1/5=3/20
=>x=3/20+1/5
=>x=3/20+4/20=7/20
c: \(\dfrac{5}{6}-x=1\)
=>x=5/6-1=-1/6
Nếu đề hỏi nhận định nào sai thì đáp án D là nhận định sai, vì theo đề thì $c$ là phần tử nằm trong tập hợp M
A. = 64.(4.9)+(32.2).137-(8.8).73
= 64.36+64.137-64.73
= 64.( 36+137-73)
= 64.100
= 6400
B. = 16.(3.13)+(2.8).150-(4.4).(62.3)
= 16.39+16.150-16.186
= 16.(39+150-186)
=16.3
= 48
C. =49.62+38.73+(31.2).51+(19.2).27
=49.62+38.73+62.51+38.27
= 62.(49+51)+38.(73+27)
= 62.100+38.100
=100.(62+38)
=100.100
=10000
a, 12- (-4)= 16
b, 12- (-14)= 26
c, (-13)-(-5)=-8
d, (-2)-(-10)= 8
\(ab+a+b=1\Leftrightarrow ab+a+b+1=2\Leftrightarrow\left(a+1\right)\left(b+1\right)=2\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)=2;\text{ }\left(b+1\right)\left(c+1\right)=-8\text{; }\left(c+1\right)\left(a+1\right)=-4\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(b+1\right)\left(c+1\right)\left(c+1\right)\left(a+1\right)=2.\left(-8\right).\left(-4\right)\)
\(\Leftrightarrow\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\right]^2=64=8^2\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)=8\text{ hoặc }\left(a+1\right)\left(b+1\right)\left(c+1\right)=-8\)
\(+\text{TH1: }\left(a+1\right)\left(b+1\right)\left(c+1\right)=8\)
\(a+1=\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(b+1\right)\left(c+1\right)}=\frac{8}{-8}=-1\Rightarrow a=-2\)
\(b+1=\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(c+1\right)\left(a+1\right)}=\frac{8}{-4}=-2\Rightarrow b=-3\)
\(c+1=\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(a+1\right)\left(b+1\right)}=\frac{8}{2}=4\Rightarrow c=3\)
\(+\text{TH2: }\left(a+1\right)\left(b+1\right)\left(c+1\right)=-8\)
Tương tự trường hợp 1; ta giải được
\(a=0;\text{ }b=1;\text{ }c=-5\)
Vậy \(\left(a,b,c\right)=\left(0;1;-5\right);\left(-2;-3;3\right)\)
a)
\(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)
\(A=-a-b+c-\left(-a\right)+b+c\)
\(A=-a+\left(-b\right)+c+a+b+c\)
\(A=\left[\left(-a\right)+a\right]+\left[\left(-b\right)+b\right]+\left(c+c\right)\)
\(A=0+0+2c\)
\(A=2c\)
____________________________________________________________________________
b)
Cách 1 : \(A=\left(-1-\left(-1\right)+\left(-2\right)\right)-\left(1-\left(-1\right)-\left(-2\right)\right)\)
\(A=-1-\left(-1\right)+\left(-2\right)-\left(-1\right)+\left(-1\right)+\left(-2\right)\)
\(A=-1+1+\left(-2\right)+1+\left(-1\right)+\left(-2\right)\)
\(A=\left[\left(-1\right)+1+1+\left(-1\right)\right]+\left[\left(-2\right)+\left(-2\right)\right]\)
\(A=0+\left(-4\right)=\left(-4\right)\)
Cách 2 : Từ ý a suy ra :
\(A=\left(-2\right)\cdot2=\left(-4\right)\)
\(a,x=-\dfrac{7}{12}-\dfrac{5}{12}=-1\)
\(b,\Rightarrow x=14-13=1\)
\(c,x+\dfrac{4}{5}=\dfrac{79}{30}\Leftrightarrow x=\dfrac{79}{30}-\dfrac{4}{5}=\dfrac{79-24}{30}=\dfrac{55}{30}=\dfrac{11}{6}\)
b \(\dfrac{x}{20}=\dfrac{27}{20}\)
\(x=\dfrac{27}{20}\)x \(20\)
\(x=27\)
c \(x+\dfrac{4}{5}=\dfrac{79}{30}\)
\(x=\dfrac{79}{30}-\dfrac{4}{5}\)
\(x=\dfrac{11}{6}\)