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a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)
\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)
Do đó: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có:
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)
Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)
áp dụng tính chất dãy tỉ số = nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)
= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)
a)đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)a=bk, c=dk
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
từ (1),(2)\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
câu c bn tự giải nhé dễ mak ahihihi
chúc bn hc tốt
a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)
Áp dụng tỉ lệ thức ta có :
\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\)\(\frac{4a}{4c}=\frac{3b}{3d}\Rightarrow\frac{4a+3b}{4c+3d}=\frac{4c-3d}{4c-3d}\)
b) Có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)
Áp dụng tỉ lệ thức ta có "
\(\frac{2a}{3b}=\frac{2c}{3d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a-3b}{2c-3d}=\frac{2a3b}{2c+3d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)
Các câu còn lại bạn làm tương tự
a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)
Áp dụng dãy tỉ số bằng nhau ta có :
\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\frac{4a-3b}{4a+3b}=\frac{4c-3d}{4c+3d}\Rightarrow\frac{4a-3d}{4c-3d}=\frac{4a+3b}{4c+3d}\)
b) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)
Áp dụng dãy tỉ số bằng nhau ta có :
\(\frac{2a}{3b}=\frac{2c}{2d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)
bạn cứ đặt công thức gốc là k sau đó thay vào các câu là được thui
*a/b=c/d=k=>a=bk;c=dk
Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3
Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3
=>2a+3b/2a-3b=2c+3d/2c-3d
*a/b=c/d=>a/c=b/d=k
=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)
k^2=a/c.b/d=ab/cd (2)
Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2
*a/b=c/d=>a/c=b/d=k=a+b/c+d
=>k^2=(a+b/c+d)^2
k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2
=>(a+b/c+d)^2=a^2+b^2/c^2+d^2
Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)
a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)
Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(2)
Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
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:
\(\dfrac{7a^2+3ab}{7c^2+3cd}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)=\(\dfrac{7a^2+3ab}{11a^2-8b^2}\)=\(\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)
\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)
=>\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
c: \(\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b}{d}\right)^4\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4}{d^4}\)
Do đó: \(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)