cho \(\dfrac{a}{b}=\dfrac{c}{d}\)(b,d≠0) chứng tỏ rằng \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a.c}{b.d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\) (1)
Lại có vì : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\) (2)
Từ (1) và (2) => ĐPCM
cho \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) cm rằng
a) \(\dfrac{a}{a-b}\) =\(\dfrac{c}{c-d}\) b)\(\dfrac{a}{b}\) =\(\dfrac{a+c}{b+d}\) c) \(\dfrac{a}{3a+d}\) =\(\dfrac{c}{3c+d}\) d)\(\dfrac{a.c}{b.d}\) =\(\dfrac{a^2+c^2}{b^2+c^2}\) e)\(\dfrac{a.b}{c.d}\) =\(\dfrac{a^2-b^2}{c^2-d^2}\) f)\(\dfrac{a.b}{c.d}\) =\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mn giúp mk vs ạ! thanks
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\),Chứng minh \(\dfrac{a^2}{b^2}=\dfrac{a.c}{b.d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk\),\(c=dk\)
\(\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\left(1\right)\)
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)(đpcm)
Đặt \(\dfrac{a}{b}=k;\dfrac{c}{d}=k\)
\(\Rightarrow a=kb;c=kd\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=k^2\)
Từ các chứng minh trên cho ta thấy
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a.c}{b.d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\)
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{k^2bd}{bd}=k^2\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\Rightarrowđpcm\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)
CMR: \(\dfrac{a.c}{b.d}\) = \(\dfrac{a^2+c^2}{b^2+d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\\ \dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh :
a) \(\dfrac{a.c}{b.d}=\dfrac{a^2-c^2}{b^2-d^2}\)
b) \(\dfrac{a^2}{b^2}=\dfrac{3a^2-2ac}{3b^2-2bd}\)
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/ \(\dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{3a-2c}{3b-2d}=\dfrac{3a^2-2ac}{3b^2-2bd}\)
Cho tỉ lệ thức \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) . Chứng minh rằng : \(\dfrac{a^2}{b^2}\) = \(\dfrac{c^2}{d^2}\) = \(\dfrac{ac}{bd}\)
Các bạn nhớ giải nhanh giúp mình nhé !
Ai làm nhanh nhất sẽ được tick 5 sao!!!
Áp dụng công thức tỉ lệ phân số ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{ac}{bd}\)
Cho các số a,b,c,d ≠ 0 và \(b^2=a.c\) ; \(c^2=b.d\) ; \(b^3+c^3+d^3\ne0\). C/m rằng :\(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
\(\left.\begin{matrix} b^2=ac\Rightarrow \dfrac{a}{b}=\dfrac{b}{c} \\c^2=bd \Rightarrow \dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right\}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)
Áp dụng t/c của DTSBN , ta có :
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\\ \Rightarrow\dfrac{a^3}{b^3}=\dfrac{a^3+b^3+c^3}{d^3+c^3+d^3}\left(1\right)\)
Có `a^3/b^3=a/b*a/b*a/b=a/b*b/c*c/d=a/d` ( do `a/b=b/c=c/d` )`(2)
Từ `(1);(2)=>` \(\dfrac{a}{d}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
cho \(b^2\)=a.c; \(c^2\)= b.d. CMR: \(\left(\dfrac{a.b.c}{b.c.d}\right)\)^2=\(\dfrac{a}{d}\)
Đề sai rồi bạn. Phải thay "^2" bằng "^3" mới đúng.
1/Cho \(\dfrac{a}{b}=\dfrac{c}{d}\left(b\ne0;d\ne0\right)\)chứng tỏ rằng\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a.c}{b.d}\)
2/Tìm x, y thỏa mãn:\(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
3/Tìm các số a, b, c biết \(\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c\) và a - b =15
4/Chứng minh M=3x+1+3x+2+3x+3+ . . . +3x+100 chia hết cho 120(x ∈ N)
Giúp mình vs mình đg gấp. Trả lời 1 câu cx đc mình sẽ tick
1. Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) \(\left(2\right)\)
Từ \(\left(1\right)\text{và (2)}\) \(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
3. \(\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)
\(\text{Mà }a-b=15\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}a=60\\b=45\\c=40\end{matrix}\right.\)
Ủng hộ bài 4 đây :V
\(M=3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(M=3^x.3^1+3^x.3^2+3^x.3^3+...+3^x.3^{100}\)
\(M=3^x\left(3^1+3^2+3^3+...+3^{100}\right)\)
Đặt: \(T=3^1+3^2+3^3+...+3^{100}\)
\(T=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(T=1\left(3^1+3^2+3^3+3^4\right)+3^4\left(3^1+3^2+3^3+3^4\right)+...+3^{96}\left(3^1+3^2+3^3+3^4\right)\)
\(T=\left(1+3^4+...3^{96}\right)\left(3^1+3^2+3^3+3^4\right)=120\left(1+3^4+...+3^{96}\right)⋮120\)
\(\Rightarrow M⋮120\left(đpcm\right)\)
