Bỏ dấu ngoặc và rút gọn biểu thức :
a) \(\left(a+b\right)\left(a+b\right)\)
b) \(\left(a-b\right)\left(a-b\right)\)
Tính giá trị biểu thức sau khi đã bỏ dấu ngoặc
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
A = (3,1 – 2,5) – (-2,5 + 3,1) = 3,1 – 2,5 + 2,5 – 3,1 = 0
B = (5,3 – 2,8) – (4 + 5,3) = 5,3 – 2,8 – 4 – 5,3
A=(3,1-2,5)-(-2,5+3,1)=3,1-2,5+2,5-3,1=0
B=(5,3-2,8)-(4+5,3)=5,3-2,8-4-5,3=-6,8
A = ( 3,1 − 2,5 ) − ( −2,5 + 3,1 )
A = 3, 1 - 2,5 + 2,5 - 3,1
A = 0
B = ( 5,3 − 2,8 ) − ( 4 + 5,3 )
B = 5,3 - 2,8 - 4 - 5,3
B = -6,8
Bỏ dấu ngoặc rồi thu gọn biểu thức :
a, \(A=\left(a-2b+c\right)-\left(a-2b-c\right)\)
b, \(B=\left(-x-y+3\right)-\left(-x+2-y\right)\)
c, \(C=2.\left(3a+b-1\right)-3\left(2a+b-2\right)\)
a) \(A=\left(a-2b+c\right)-\left(a-2b-c\right)\)
\(A=a-2b+c-a+2b+c=2c\)
b) \(B=\left(-x-y+3\right)-\left(-x+2-y\right)\)
\(B=-x-y+3+x-2+y=1\)
c) \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)\)
\(C=6a+2b-2-6a-3b+6=4-b\)
a. \(A=\left(a-2b+c\right)-\left(a-2b-c\right)=a-2b+c-a+2b+c=0\)
b. \(B=\left(-x-y+3\right)-\left(-x+2-y\right)=-x-y+3+x-2+y=1\)
c. \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)=6a+2b-2-6b-3b+6=4-3b\)
Sử dụng công thức cộng, rút gọn mỗi biểu thức sau:
\(\cos \left( {a + b} \right) + \cos \left( {a - b} \right);\,\,\cos \left( {a + b} \right) - \cos \left( {a - b} \right);\,\,\sin \left( {a + b} \right) + \sin \left( {a - b} \right)\)
\(\begin{array}{l}\cos \left( {a + b} \right) + \cos \left( {a - b} \right) = \cos a.\cos b - \sin a.\sin b + \sin a.\sin b + \cos a.\cos b = 2\cos a.\cos b\\\cos \left( {a + b} \right) - \cos \left( {a - b} \right) = \cos a.\cos b - \sin a.\sin b - \sin a.\sin b - \cos a.\cos b = - 2\sin a.\sin b\\\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = \sin a.\cos b + \cos a.\sin b + \sin a.\cos b - \cos a.\sin b = 2\sin a.\cos b\end{array}\)
Rút gọn biểu thức :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)
Rút gọn biểu thức :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b^2-cb-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-\left(b+a-c\right)}{ab\left(a-c\right)\left(b-c\right)}\)
Rút gọn biểu thức :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(-a-b+c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=-\dfrac{a+b-c}{ab\left(b-c\right)\left(a-c\right)}\)
Rút gọn biểu thức: \(A=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}+\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
Rút gọn biểu thức\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Rút gọn biểu thức
a. B = \(\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
b. C = \(a:\left(b-2\right)-\left[\left(a^2+2a+1\right):\left(b^2-4\right)\right].\left[\left(b+2\right):\left(a+1\right)\right]\)