Rút gọn biểu thức sau: \(\frac{\left(a.b+b.c+c.d+d.a\right).a.b.c.d}{\left(c+d\right).\left(a+b\right)+\left(b-c\right).\left(a-d\right)}\)
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Rút gọn biểu thức sau:
\(\frac{\left(a.b+b.c+c.d+d.a\right).a.b.c.d}{\left(c+d\right).\left(a+b\right)+\left(b-c\right).\left(a-d\right)}\)
1/ cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng:
a) \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\frac{a.d}{c.b}=\frac{\left(a+b\right).\left(a-b\right)}{\left(c+d\right).\left(c-d\right)}\)
2/ cho a.b=c2 chứng minh: \(\frac{a}{b}=\frac{\left(2.a+3.c\right)^2}{\left(2.c\right)+\left(3.b\right)^2}\)
1/ cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a) \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b)\(\frac{a,d}{c.b}=\frac{\left(a+b\right).\left(a-b\right)}{\left(c+d\right).\left(c-d\right)}\)
2/ cho \(a.b=c^2\)chứng minh : \(\frac{a}{b}=\frac{\left(2a+3c\right)^2}{\left(2c+3b\right)^2}\)
a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)
Thay:
\(\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
=> đpcm
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Cho\(\frac{a}{b}=\frac{c}{d};\left(a,b,c,d,\right)\)chứng minh\(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Vì \(\frac{a}{b}=\frac{c}{d}\) nên ad=bc và \(\frac{a}{c}=\frac{b}{d}=\frac{ab}{cd}\)(1)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{c+d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(2)
Từ (1) và (2), ta suy ra: \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
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Rút gọn các Biểu Thức sau
a)\(\frac{a^2m-a^2n-b^2n+b^2m}{a^2+b^2}\)
b)\(\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}\)
a) \(\frac{a^2m-a^2n-b^2n+b^2m}{a^2+b^2}=\frac{a^2\left(m-n\right)+b^2\left(m-n\right)}{a^2+b^2}\)
\(=\frac{\left(m-n\right)\left(a^2+b^2\right)}{a^2+b^2}=m-n\)
b) \(\frac{\left(ab+bc+cd+ad\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-b\right)}\)
\(=\frac{\left[b.\left(a+c\right)+d.\left(a+c\right)\right].abcd}{ac+bc+da+db+ab-b^2-ca+bc}\)
\(=\frac{\left(a+c\right)\left(d+b\right)abcd}{2bc+da+db+ab-b^2}\)
Rút gọn biểu thức sau :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
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cho a/b = c/d . tính \(\frac{a.b}{c.d}+\left[\left(\frac{a+b}{c+d}\right)^2:\left(\frac{a^2+b^2}{c^2+d^2}\right)\right]-\frac{a^2-b^2}{c^2-d^2}\)
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\left(\frac{a+b}{c+d}\right)^2=\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\)
=> \(\frac{ab}{cd}+\left[\left(\frac{a+b}{c+d}\right)^2:\left(\frac{a^2+b^2}{c^2+d^2}\right)\right]-\frac{a^2-b^2}{c^2-d^2}\)
= \(\frac{ab}{cd}+1-\frac{a^2-b^2}{c^2-d^2}\)
= \(1\)
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Rút gọn biểu thức :
\(\frac{a^2\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{b^2\left(b+a\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{c^2\left(c+a\right)\left(c+b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)