ĐƠN GIẢN CÁC BIỂU THỨC
\(a,-b-\left(b-a+c\right)\) \(b,-\left(a-b+c\right)-\left(c-a\right)\)
\(c,b-\left(b+a-c\right)\) \(d,a-\left(-b+a-c\right)\)
Đơn giản biểu thức
\(A=\frac{\left(a-2\right)\left(a-1014\right)}{a\left(a-b\right)\left(a-c\right)}+\frac{\left(b-2\right)\left(b-1004\right)}{b\left(b-a\right)\left(b-c\right)}+\frac{\left(c-2\right)\left(c-1004\right)}{c\left(c-a\right)\left(c-b\right)}\)
mn giúp mik vs
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))
Đơn giản biểu thức sau :
\(N=\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a}+\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(N=\dfrac{\left(a-b\right)\left(b+c\right)\left(a+c\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)+\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(ab-b^2+ac-bc+ab-ac+b^2-cb\right)+\left(c-a\right)\left(ab+b^2+ac+bc+ab-b^2-ac+cb\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(2ab-2bc\right)+\left(c-a\right)\left(2ab+2bc\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{2b\left(a+c\right)\left(a-c\right)+2b\left(c-a\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2b\left(c+a\right)\left(a-c+c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
Mình chỉ biết mỗi cách quy đồng...... Rồi kết hợp ....
Đơn giản biểu thức:
\(A=\left(a-b+c\right)-\left(b-c-d\right)+\left(c-d+a\right)\)
\(B=\left(a+b-c\right)+\left(b+c-a\right)-\left(a-c\right)\)
\(C=-\left(4a+5b-c\right)-\left(5b+3c\right)\)
\(D=\left(a-3b+c\right)-\left(2a-b+c\right)\)
A=(a-b+c)-(b-c-d)+(c-d+a)
A=a-b+c-b+c+d+c-d+a
A=2a-2b-3c
B=( a + b - c ) + ( b + c - a ) - ( a - c )
B=a + b - c + b + c - a - a + c
B=2b + c - a
C = - ( 4a + 5b + c) - ( 5b + 3c )
C = -4a - 5b - c - 5b -3c
C= -4a - 10b - 4c
D= ( a - 3b + c) - ( 2a -b +c)
D= a - 3b +c - 2a + b -c
D= a - 2b
Rút gọn các biểu thức:
a) \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2\)
b) \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
c ) \(\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
Rút gọn các biểu thức
\(a.\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(b.\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
a. \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2-2ab+2bc-2ac+c^2+a^2+b^2-2bc+2ac-2ab+a^2+b^2+c^2+2ab-2ac-2bc=4\left(a^2+b^2+c^2\right)\)b. Bạn làm tương tự câu a , đáp số ra : \(4\left(a^2+b^2+c^2+d^2\right)\)
Đơn giản biểu thức bằng vận dụng tính chất nghiệm đa thức:
N = \(\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Gọi x1 là nghiệm âm của phương trình: x2 + x -1 =0.
Không giải phương trình tính giá trị của:
\(D=\sqrt{x_1^8+10x_1+13}+x_1\)
Rút gọn biểu thức :
a . \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
b . \(\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
Rút gọn biểu thức :
\(\frac{a^2\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{b^2\left(b+a\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{c^2\left(c+a\right)\left(c+b\right)}{\left(c-a\right)\left(c-b\right)}\)
Tính giá trị của biểu thức với a,b,c khác nhau:
\(M=\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}.\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}.\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\)
\(M=\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}.\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}.\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\) (loạn mắt quá !!! )
\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c-a\right)\left(b+c\right)\left(c+a\right)\left(a-b\right)\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Thôi để bữa sau a làm (đau mắt quá :)))