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\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)
\(a,\) Áp dụng tcdtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)
\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3b}{3d}=\dfrac{5a}{5c}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\\ \Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{5a}{3b}=\dfrac{5c}{3d}\)
hay \(\dfrac{5a}{5c}=\dfrac{3b}{3d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Leftrightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
hay \(\dfrac{5a+3n}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)(đpcm)
Mk chỉ làm 1 câu thôi mấy câu sau tương tự theo cách đó nhoa:v
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^4=\dfrac{b^4}{d^4}\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{bk^4+b^4}{dk^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\Rightarrowđpcm\)
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^4}{c^4}\)=\(\dfrac{b^4}{d^4}\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^4\)(2)
Từ (1) và (2)suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^4\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(đpcm)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a+3b}{5c+3d}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5b}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a-3b}{5c-3d}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)=\(\dfrac{5a+3b}{5a-3b}\)=\(\dfrac{5c+3d}{5c-3d}\) (đpcm)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Do đó: \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{a}{c}\right)^2\)và \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{b}{d}\right)^2\)
=>\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\) và \(\dfrac{ab}{cd}\)=\(\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{7a^2}{7c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{3ab}{3cd}\)=\(\dfrac{7a^2+3ab}{7c^2+3cd}\)(1)
Ta có: \(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=> \(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{7a^2+3ab}{7c^2+3cd}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)=\(\dfrac{7a^2+3ab}{11a^2-8b^2}\)=\(\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Lời giải:
Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk$.
Khi đó:
$\frac{5a+3b}{5a-3b}=\frac{5bk+3bk}{5bk-3bk}=\frac{8bk}{2bk}=4(1)$
$\frac{5c+3d}{5c-3d}=\frac{5dk+3dk}{5dk-3dk}=\frac{8dk}{2dk}=4(2)$
Từ $(1); (2)$ suy ra điều phải chứng minh.
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
Do đó: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b: \(\dfrac{7a^2+8ab}{11a^2-8b^2}=\dfrac{7b^2k^2+8\cdot bk\cdot b}{11\cdot b^2\cdot k^2-8b^2}=\dfrac{7k^2+8k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+8\cdot dk\cdot d}{11\cdot d^2\cdot k^2-8d^2}=\dfrac{7k^2+8k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2+8ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\) (1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\) (2)
Từ (1) và (2) suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) Từ (*) ta có:
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\) (3)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\) (4)
Từ (3) và (4) suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
P/s: test lại đề phần b), mẫu số của vế trái
a, Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Vì \(\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}\)\(=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\left(đpcm\right)\)
Vậy \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+5d}{5c-5d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
a, Ta có: \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)\(\Rightarrow\dfrac{\left(bk-b\right)^2}{\left(ck-c\right)^2}=\dfrac{bk.b}{dk.d}\)
\(\Rightarrow\dfrac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\)
Vậy \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)
b, Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}\)
\(\Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a+3b}{5c+3d}\)
Bài 2:
\(A=\frac{8^5(-5)^8+(-2)^5.10^9}{2^{16}.5^7+20^8}\) \(=\frac{(2^3)^5(-5)^8+(-2)^5.2^9.5^9}{2^{16}.5^7+(2^2.5)^8}\)
\(=\frac{2^{15}.5^8-2^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)
\(=\frac{5(-3)}{2^2.6}=\frac{-5}{8}\)
Bài 3:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Thay vào:
\(\frac{5a+3b}{5a-3b}=\frac{5bt+3b}{5bt-3b}=\frac{b(5t+3)}{b(5t-3)}=\frac{5t+3}{5t-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dt+3d}{5dt-3d}=\frac{d(5t+3)}{d(5t-3)}=\frac{5t+3}{5t-3}\)
Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
Bài 4:
Ta có:
\(A=3+3^2+3^3+3^4+...+3^{100}\)
\(=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{97}+3^{98}+3^{99}+3^{100})\)
\(=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+...+3^{97}(1+3+3^2+3^3)\)
\(=3.40+3^5.40+....+3^{97}.40\)
\(=120(1+3^4+....+3^{96})\vdots 120\)
Ta có đpcm.