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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)
a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)\(=\dfrac{\dfrac{a}{k}.b}{\dfrac{c}{k}.d}=\dfrac{ab}{cd}=VT\)
Vậy...
b) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
Suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(bk\right)^2+3\left(bk\right).b}{11\left(bk\right)^2-8b^2}\)\(=\dfrac{7k^2+3k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3\left(dk\right).d}{11\left(dk\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)
Suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
a) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(ad=bc\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\) => \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
(theo tính chất dãy tỉ số bằng nhau)
=> (đpcm)
b) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)(theo tính chất dãy tỉ số bằng nhau)
=> \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\) (đpcm)
c) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{a^2}{c^2}=\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\) => \(\dfrac{7a^2}{7c^2}=\dfrac{3ab}{3cd}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}\)
=> \(\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\) (theo tính chất dãy tỉ số bằng nhau)
=> \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)(đpcm)
#Ayumu
a) dk: \(\left\{{}\begin{matrix}a,d\ne0\\5a\ne3b\\5c\ne3d\end{matrix}\right.\) \(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5.\dfrac{a}{b}+3}{5\dfrac{a}{b}-3}=\dfrac{5.\dfrac{c}{d}+3}{5\dfrac{c}{d}-3}=\dfrac{\dfrac{5c+3d}{d}}{\dfrac{5c-3d}{d}}=\dfrac{5c+3d}{d}.\dfrac{d}{5c-3d}=\dfrac{5c+3d}{5c-3d}=VP\)
b)
\(\left\{{}\begin{matrix}b,d\ne0\\11a^2\ne8b^2\\11c^2\ne8d^2\end{matrix}\right.\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\right)\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7.\dfrac{a^2}{b^2}+3\dfrac{a}{b}}{11\dfrac{.a^2}{b^2}-8}=\dfrac{7.\dfrac{c^2}{d^2}+3\dfrac{c}{d}}{11\dfrac{.c^2}{d^2}-8}=\dfrac{7c^2+3cd}{11c^2-8d^2}=VP\)
a) \(\dfrac{a}{b}=\dfrac{c}{d}\left(a;b;c;d\ne0\right)\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\Rightarrow dpcm\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
\(\Rightarrow dpcm\)
c) 22/5 + 51/9 + 11/4 + 3/5 + 1/3 + 1/4
= 22/5 +3/5 +51/9 + 1/3 +11/4+1/4
= (22/5 +3/5) +(51/9 + 3/9) +(11/4+1/4)
= 25/5 +54/9 +12/4
= 5 +6 +3
= 14
d) (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15)
= (5/30 + 3/30 +2/30 ) :(5/30 +3/30 -2/30)
= 10/30 : 6/30
= 1/3 : 1/5
= 5/3
Cảm ơn pn Bexiu ^^ Nhưng đây là c/m mà bn ;) ;) Có phải tính đâu =)) Nhưng ko sao ah :3 Cảm ơn pn đã giúp <3
Đặt PT đã cho ở đề là A
Ta có : \(\sqrt{3a^2+8b^2+14ab}=\sqrt{3a\left(a+4b\right)+2b\left(a+4b\right)}=\sqrt{\left(3a+2b\right)\left(a+4b\right)}\)
\(\le\dfrac{3a+2b+a+4b}{2}=\dfrac{4a+6b}{2}=2a+3b\)
\(\Rightarrow\dfrac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\dfrac{a^2}{2a+3b}\)
Làm tương tự như trên , ta có :
\(\dfrac{b^2}{\sqrt{3b^2+8c^2+14bc}}\ge\dfrac{b^2}{2b+3c};\dfrac{c^2}{\sqrt{3c^2+8a^2+14ac}}\ge\dfrac{c^2}{2c+3a}\)
Nên : \(A\ge\dfrac{a^2}{2a+3b}+\dfrac{b^2}{2b+3c}+\dfrac{c^2}{2c+3a}\ge\dfrac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\dfrac{5}{a+b+c}\left(đpcm\right)\)
Đây là câu a/
https://hoc24.vn/hoi-dap/question/693692.html?pos=1903228
Còn câu b thì như sau:
Trước hết, nghi ngờ bạn ghi sai đề ở con này \(\dfrac{1}{a^2+7a+9}\) , số 9 phải là số 12 mới hợp lý. Mình tự sửa lại đề, còn nếu đề đúng như bạn chép thì bạn giữ nguyên nó, phần còn lại rút gọn được còn đâu thì quy đồng giải trâu thôi, chẳng cách nào với đề xấu kiểu ấy cả.
\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a-2c}{3b-2d}\)
a/ \(\dfrac{a.c}{b.d}=\dfrac{\left(a+c\right).\left(a-c\right)}{\left(b+d\right).\left(b-d\right)}=\dfrac{a^2-c^2}{b^2-d^2}\)
b/ \(\dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{3a-2c}{3b-2d}=\dfrac{3a^2-2ac}{3b^2-2bd}\)
kcjHoàng Oanh
1.\(\dfrac{5a+3b}{5a-3b}\)=\(\dfrac{5c+3d}{5c-3d}\)
Ta có:\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}\)
=>\(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)(a/d t/c của dãy tỉ số bằng nhau)
=>\(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)(đpcm)