Cho \(\frac{a}{b}<\frac{c}{d}\) và b,d >0. Chứng minh rằng: \(\frac{a}{b}>\frac{ab+cd}{b^2+d^2}<\frac{c}{d}\)
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Cho a/b < c/d và b,d>0. Chứng minh rằng: \(\frac{a}{b}
1. a)Cho a-b+c-d=0. Chứng minh rằng: a3 - b3 + c3 - d3=3(c-d)(cd-ab)
b) cho a+d=b-c. Chứng minh rằng: a3 - b3 + c3 + d3=3(a-b)(ab+dc)
2. a)Cho \(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\)=0. Tính S= \(\frac{yz}{x^2}-\frac{xy}{z^2}-\frac{zx}{y^2}\)
b) Cho \(\frac{1}{x}+\frac{2}{y}+\frac{3}{z}\)=0. Tính S= \(\frac{9xy}{2z^2}+\frac{yz}{6x^2}+\frac{4zx}{3y^2}\)
Cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)với a,b,c,d khác 0 ; c khác +d và -d . chứng minh rằng hoặc a/b = c/d hoặc a/b = d/c
Ta có :
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}-\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{b}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\text{ hay }\frac{a}{b}=\frac{c}{d}\)
TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)
Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\text{ hay }\frac{a}{b}=\frac{d}{c}\)
Vậy : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\text{ thì }\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
kinh quá
Cho \(a,b,c,d>0\). Chứng minh rằng: \(\frac{1}{a^2+ab}+\frac{1}{b^2+bc}+\frac{1}{c^2+cd}+\frac{1}{d^2+da}\ge\frac{4}{ac+bd}\)
Svacxo chăng :33 Ai thử đi, e sợ biến nhiều lắm :))
Biết \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)với a,b,c,d khác 0. Chứng minh rằng: \(\frac{a}{b}=\frac{c}{d}\)hoặc\(\frac{a}{b}=\frac{d}{c}\)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) => \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
TH2: \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b+b-a}{c+d+d-c}=\frac{2b}{2d}=\frac{b}{d}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b-b+a}{c+d-d+c}=\frac{2a}{2c}=\frac{a}{c}\left(6\right)\)
Từ (5) và (6) => \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
\(\frac{b}{c}=\frac{a}{d}\)ở đâu vậy
Cho tỉ lệ thức \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) với a,b,c,d khác 0 và c khác-d
Chứng minh rằng : \(\frac{a}{b}=\frac{c}{d}\)hoặc \(\frac{a}{b}=\frac{d}{c}\)
Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow\left(a^2cd-abd^2\right)+\left(b^2cd-abc^2\right)=0\)
\(\Leftrightarrow ad\left(ac-bd\right)-bc\left(ac-bd\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (đpcm)
Cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)với a,b,c khác 0;\(c\ne\pm d\).chứng minh rằng hoặc \(\frac{a}{b}=\frac{d}{c}\)hoặc
Ta có\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
<=> cd(a2 + b2) = ab(c2 + d2)
<=> a2cd + b2cd - abc2 - abd2 = 0
<=> (a2cd - abc2) + (b2cd - abd2) = 0
<=> ac(ad - bc) + bd(bc - ad) = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> \(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{a}{d}=\frac{b}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\left(\text{đpcm}\right)\)
cho\(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng
\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
Đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)= k ( k \(\in\)Z , k khác 0 )
=> a = bk ; c = dk
Ta có:
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
Vậy nếu \(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
Cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\).Chứng minh rằng: \(\frac{a}{b}=\frac{c}{d}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right).cd=ab.\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abd^2=abc^2-b^2cd\)
\(\Leftrightarrow ad\left(ac-bd\right)=bc\left(ac-bd\right)\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
Ta co:a^2+b^2•cd=c^2+d^2•ab=>(a+b)^2•ab=(c+d)^2•cd=>(a+b)^3=(c+d)^3=>a•(b^3)=c•(d^3)=>a/c=b^3/d^3=>a/c=b/d=>a/b=c/d. Do la dieu Phai Chung minh