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Khi a=-4/5
= > A=-4/5.1/2+(-4/5).1/3+(-4/5).1/4
A=-4/5.(1/2+1/3-1/4)
A=-4/5.7/12
A=-7/15
Các bài còn lai tương tự
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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Bài 3 :
a) \(1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right)\)\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[19+\left(-20\right)\right]\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\cdot10=-10\)
b) \(1-2+3-4+...+99-100=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\cdot50=-50\)
c) \(2-4+6-8+...+46-48+50-52=\left(2-4\right)+\left(6-8\right)+...+\left(50-52\right)\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)\)
\(=\left(-2\right)\cdot13=-26\)
d) \(-1+3-5+7-...-97+99\)\(=\left(-1+3\right)+\left(-5+7\right)+...+\left(-97+99\right)\)
\(=2+2+...+2\)
\(=2\cdot25=50\)
e) \(1+\left(-2\right)+3+\left(-4\right)+...+1999+\left(-2000\right)+2001\)\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[1999+\left(-2000\right)\right]+2001\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2001\)
\(=\left(-1\right)\cdot1000+2001=-1000+2001=1001\)
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Bài 4 :
a) \(\left(2ab^2\right):\left(abc\right)=\left[2\cdot4\cdot\left(-6^2\right)\right]:\left[4\cdot\left(-6\right)\cdot12\right]\)
\(=\left[2\cdot4\cdot36\right]:\left[4\cdot\left(-6\right)\cdot12\right]\)
\(=\left[8\cdot36\right]:\left[-24\cdot12\right]\)
\(=288:\left(-288\right)=-1\)
b) \(\left[\left(-25\right)\cdot\left(-27\right)\cdot\left(-x\right)\right]:y=\left[\left(-25\right)\cdot\left(-27\right)\cdot4\right]:\left(-9\right)\)
\(=\left[675\cdot4\right]:\left(-9\right)=2700:\left(-9\right)=-300\)
c) \(\left(a^2-b^2\right):\left(a+b\right)\left(a-b\right)=\left(5^2-\left(-3^2\right)\right):\left(5+\left(-3\right)\right)\left(5-\left(-3\right)\right)\)
\(=\left(25-9\right):\left(5+\left(-3\right)\right)\left(5-\left(-3\right)\right)\)
\(=16:2\cdot8=8\cdot8=64\)
a) \(\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\)
=> \(\frac{12}{13}x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)
=> \(x=2:\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)
b) \(x:\frac{13}{3}=-2,5\)
=> \(x:\frac{13}{3}=-\frac{5}{2}\)
=> \(x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)
c) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)
=> \(\frac{4x-3}{12}=-\frac{10}{12}\)
=> 4x - 3 = -10
=> 4x = -10 + 3 = -7
=> x = -7/4
Bài 2 :
\(A=a\cdot\frac{1}{3}+a\cdot\frac{1}{4}-a\cdot\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\cdot\frac{5}{12}\)
Thay a = -3/5 vào biểu thức ta có : \(A=\left(-\frac{3}{5}\right)\cdot\frac{5}{12}=\frac{-3}{12}=\frac{-1}{4}\)
\(B=b\cdot\frac{5}{6}+b\cdot\frac{3}{4}-b\cdot\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\cdot\frac{13}{12}\)
Thay b = 12/13 vào ta được kết quả là 1
a ) \(\frac{25}{9}-\frac{12}{13}\cdot x=\frac{7}{9}\)
\(\Rightarrow\frac{12}{13}\cdot x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)
\(\Rightarrow x=2\div\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)
Vậy ...
b ) \(x\div\frac{13}{3}=-\frac{5}{2}\)
\(\Rightarrow x\div\frac{13}{3}=-\frac{5}{2}\)
\(\Rightarrow x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)
Vậy ..
c ) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)
\(\Rightarrow\frac{4x-3}{12}=-\frac{10}{12}\)
\(\Rightarrow4x-3=-10\)
\(\Rightarrow4x=-10+3=-7\)
\(\Rightarrow x=-\frac{7}{4}\)
Vậy ....
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
1:a.(2ab^2):c=(2.4.-6):12
=(-48):12
= - 4
b.[(-25).(-27).(-x)]:y= [(-25).(-27).(-4)]:-9
= (-2700): -9
= 300
c.(a^2 - b^2):(a+b).(a-b) = (5^2 - (-3)^2):(5+(-3)).(5 - (-3)
= 64
Mình tl k mình nha
A= a .1/2+a .1/3-a .1/4 với a=-4/5
A=a.(1/2+1/3-1/4)
A=-4/5.(6/12+4/12-3/12)
A=-4/5 . 7/12
A=\(\frac{-7}{15}\)