cho\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)
Tính : M=\(\frac{2\times a-b}{c+d}+\frac{2\times b-c}{d+a}+\frac{2\times c-d}{a+b}+\frac{2\times d-a}{b+c}\)
Cho\(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
b)\(\frac{a\times c}{b\times d}=\frac{a^2+c^2}{b^2+d^2}\)
c)\(\frac{a\times b}{c\times d}=\frac{a^2-b^2}{c^2-d^2}\)
Biết \(\frac{a^2+b^2}{c^2+d^2}=\frac{a\times b}{c\times d}\) với a,b,c,d \(\ne\)0. CM: \(\frac{a}{b}=\frac{c}{d}\)hoặc\(\frac{a}{b}=\frac{d}{c}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng:
\(\frac{a^2+c^2}{b^2+d^2}=\frac{a\times c}{b\times d}\)
\(Cho:\frac{a}{b}=\frac{c}{d}.CMR:\frac{2017\times a-b}{a}=\frac{2017\times c-d}{c}\)
Ta có : \(\frac{a}{b}=\frac{c}{d}\)
Suy ra : \(\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2017a-b}{2017c-d}\)
Nên : \(\frac{a}{c}=\frac{2017a-b}{2017c-d}\)
Do đó : \(\frac{2017a-b}{a}=\frac{2017c-d}{c}\) (đpcm)
\(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{2017a}{2017c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :
\(\frac{2017a}{2017c}=\frac{b}{d}=\frac{2017a-b}{2017c-d}\)
\(\Rightarrow\frac{2017a-b}{2017c-d}=\frac{b}{d}=\frac{a}{c}\)
\(\Rightarrow\frac{2017a-b}{2017c-d}=\frac{a}{c}\)
\(\Rightarrow\frac{2017a-b}{a}=\frac{2017c-d}{c}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có:\(\frac{2017a-b}{a}=\frac{2017bk-b}{bk}=\frac{b\left(2017k-1\right)}{bk}=\frac{2017k-1}{k}\left(1\right)\)
\(\frac{2017c-d}{c}=\frac{2017dk-d}{dk}=\frac{d\left(2017k-1\right)}{dk}=\frac{2017k-1}{k}\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{2017a-b}{a}=\frac{2017c-d}{c}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)Chứng minh rằng:
a,\(\frac{a}{a+b}=\frac{c}{c+d}\)
b,\(\left(\frac{a-b}{c-d}\right)^2=\frac{a\times b}{c\times d}\)
c,\(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\)
a, Ta co : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\)\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(1)
Xet :\(\frac{a}{a+b}=\frac{c}{c+d}\Rightarrow\frac{a}{c}=\frac{a+b}{c+d}\)(2)
Tu (1) va (2) \(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
b
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng:
\(\frac{a^2+c^2}{b^2+d^2}=\frac{a\times c}{b\times d}\)
đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
\(\frac{a.c}{b.d}=\frac{bk.dk}{b.d}=k^2\)
suy ra: \(\frac{a^2+c^2}{b^2+d^2}=\frac{a.c}{b.d}\)( cùng bằng k2)
Bài 43Cho A=\(\left(\frac{1}{2^2}-1\right)\times\left(\frac{1}{3^2}-1\right)\times\left(\frac{1}{4^2}-1\right)\times....\times\left(\frac{1}{100^2}-1\right)\)
So sánh A với \(-\frac{1}{2}\)
Bài 58.Cho tỉ lệ thức \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\).Chứng minh rằng a=c hoặc a+b+c+d=0
a,\(\sqrt{49}\)-\(\sqrt{25}\)-\(\sqrt{100}\)
b, \(\left(\frac{3}{5}\right)^4\times\left(\frac{5}{3}\right)^5\)
2.Tìm x, biết :
a,\(\orbr{ }x-\frac{1}{2}\text{]}\text{ }\)\(=9,6\)
b,\(7^{2\times x}+7^{2\times x+2}=2450\)
3.Cho 4 tỉ số bằng nhau:
\(\frac{a}{b+c+d};\frac{b}{c+d+a};\frac{c}{d+a+b};\frac{d}{a+b+c}\)
Tính giá trị của mỗi tỉ số \(\left(a,b,c,d\ne0\right)\)
Cho \(\frac{a+b}{a+c}=\frac{a-b}{a-c}\) với \(a\ne c;a\ne-c;a\times c\ne0\)
tính giá trị biểu thức:
A=\(\frac{10\times b^2+9\times b\times c+c^2}{2\times b^2+b\times c+2\times c^2}\)
Dễ mà bạn!
Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b+a-b}{a+c+a-c}=\frac{2a}{2a}=1\)
\(\Rightarrow\hept{\begin{cases}a+b=a+c\\a-b=a-c\end{cases}\Leftrightarrow}b=c\)
Ta có: \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{20b^2}{5b^2}=\frac{20}{5}=4\)
Cách khác:
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}\Leftrightarrow\left(a+b\right).\left(a-c\right)=\left(a-b\right).\left(a+c\right)\)
\(\Leftrightarrow a^2-ac+ab-bc=a^2+ac-ab-bc\Leftrightarrow-ac+ab=ac-ab\Rightarrow2ac=2ab\Rightarrow b=c\)(vì a.c khác 0)
\(A=\frac{10.c^2+9c^2+c^2}{2c^2+c^2+2c^2}=\frac{20c^2}{5c^2}=4\)