2a^2-2ab/ac+ad-bc-bd rut gon phan thuc
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cho bieu thuc a^3+2a^2-1/a^3+2a^2+2a+1
a) Rut gon bieu thuc
b)Chung minh rang neu a la so nguyen thi gia tri tim duoc cua bieu thuc o cau a) la 1 phan so toi gian
cho biểu thức A= a^3 +2a^2-1/a^3 +2a^2+2a+1
a, rut gon bieu thuc
b, CMR neu A la so nguyen thi gia tri cua bieu thuc tim duoc cua cau a la 1 phan so toi gian
\(^{\dfrac{2a^2-2ab}{ac+ad-bc-bd}}\)
\(=\dfrac{2a\left(a-b\right)}{\left(a-b\right)\left(c+d\right)}=\dfrac{2a}{c+d}\)
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cho biểu thức A= a^3 +2a^2-1/a^3 +2a^2+2a+1
a, rut gon bieu thuc
b, CMR neu A la so nguyen thi gia tri cua bieu thuc tim duoc cua cau a la 1 phan so toi gian
cho bieu thuc A =\(\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
a)rut gon bieu thuc
b)chung minh rang neu a la so nguyen thi gia tri cua bieu thuc tim duoc o cau a la mot phan so toi gian
\(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
a/ Rut gon bieu thuc
b/ Chung minh rang neu a la so nguyen thi gia tri cua bieu thuc tim duoc o cau a, la mot phan so toi gian
rut gon bieu thuc
(a+b-c)^2 - (a-c)^2- 2ab+ 2bc
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=\left(a-c\right)^2+2b\left(a-c\right)+b^2-\left(a-c\right)^2-2ab+2bc\)
\(=2b\left(a-c\right)+b^2-2ab+2bc\)
\(=2ab-2bc+b^2-2ab+2bc=b^2\)
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Cho bieu thuc A=a3+2a3-1/a3+2a3+2a+1
a ,Rut gon bieu thuc
b,CMR neu a la so nguyen thi gia tri cua bieu thuc tim duoc o cau a la mot phan so toi gian.
a) \(A=\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}=\frac{a^2.\left(a+1\right)+\left(a+1\right).\left(a+1\right)}{a^2.\left(a+1\right)+a.\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right).\left(a^2+a-1\right)}{\left(a+1\right).\left(a^2+a+1\right)}=\frac{ }{ }\)\(\frac{a^2+a-1}{a^2+a-1}\)
duyệt đi
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\(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
a)rut gon bieu thuc
b)chung minh rang neu a la so nguyen thi gia tri cua bieu thuc tim duoc cua cau a la mot phan so toi gian
\(a,A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
\(=\frac{a^2\left(a+1\right)+\left(a-1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}\)
\(=\frac{\left(a+1\right)\left(a^2+a-1\right)}{\left(a+1\right)\left(a^2+a+1\right)}\)
\(=\frac{a^2+a-1}{a^2+a+1}\)
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