A. 2
B. 3
C. 3
D. 4
Những câu hỏi liên quan
2.a/3.b + 3.b/4.c + 4.c/5.d + 5.d/2.a biết 2.a/3.b = 3.b/4.c = 4.c/5.d = 5.d/2.a
A. 1 - b, 2 - a, 3 - d, 4 - c.
B. 1 - b, 2 - d, 3 - a, 4 - c.
C. 1 - c, 2 - a, 3 - d, 4 - b.
D. 1 - c, 2 - b, 3 - d, 4 - a.
A. 1-c; 2-a, d; 3-g; 4-b, e.
B. 1-c; 2-a, e; 3-d, g; 4-b.
C. 1-a, d; 2-c; 3-b, e; 4-g.
D. 1-a, e; 2-c, d; 3-b; 4-g.
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Nối cột A tương ứng với cột b
A. 1-b,2-a,3-d,4-c.
B. 1-a,2-b,3-c,4-d.
C. 1-d,2-c,3-b,4-a.
D. 1-d,2-a,3-c,4-b.
17 tháng 8 2021 lúc 20:04
2.a/3.b + 3.b/4.c + 4.c/5.d + 5.d/2.a biết 2.a/3.b = 3.b/4.c = 4.c/5.d = 5.d/2.a
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)
\(\Rightarrow\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4\)
Cho dfrac{a}{b}dfrac{c}{d} . Chứng minh :
a, dfrac{a^3+b^3}{c^3+d^3} dfrac{a^3-b^3}{c^3-d^3}
b, dfrac{(a+b)^3}{(c+d)^3}dfrac{a^3+b^3}{c^3+d^3}
c, dfrac{(a-b)^3}{(c-d)^3}dfrac{3a^2+2b^2}{3c^2+2d^2}
d, dfrac{4a^4+5b^4}{4c^4+5d^4}dfrac{a^2b^2}{c^2d^2}
e, dfrac{a^{10}+b^{10}}{(a+b)^{10}} dfrac{c^{10}+d^{10}}{(c+d)^{10}}
Đọc tiếp
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{a^3+b^3}{c^3+d^3} = \dfrac{a^3-b^3}{c^3-d^3}\)
b, \(\dfrac{(a+b)^3}{(c+d)^3}=\dfrac{a^3+b^3}{c^3+d^3}\)
c, \(\dfrac{(a-b)^3}{(c-d)^3}=\dfrac{3a^2+2b^2}{3c^2+2d^2}\)
d, \(\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)
e, \(\dfrac{a^{10}+b^{10}}{(a+b)^{10}} = \dfrac{c^{10}+d^{10}}{(c+d)^{10}}\)
a/
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}\)
Áp dụng tỉ lệ thức ta có:
\(\frac{a^3}{b^3}=\frac{c^3}{d^3}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)
Vậy \(\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)
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1. Cho tỉ lệ thức a/b=c/d chứng minh rằng
a. 2006*(a+c)/2006*a=b+d/b
b.a-b/a+b=c-d/c+d
c.2*a+5*b/3*a-4*b=2*c+5*d/3*c-4*d
d. (a+b/c+d)^3=a^3-b^3/c^3-d^3
Cho tỉ lệ thức a/b = c/d. Chứng yor rằng: 1) a/a+b = c/c+d; 2) 2.a+b/a-2.b = 2.c+d/c-2.d; 3) a+b/a-c = c+d=c-d; 4) 5.a+3.b/5.c+3.d = 5.a-3.b/5.c-3.d
1: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=b\cdot k;c=d\cdot k\)
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)
Do đó: \(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
3: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\cdot\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)
\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
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\(\dfrac{a}{3}+b=15\)
\(b-c=\dfrac{a}{4}\)
\(\dfrac{c}{4}+b=d\)
\(a+b+c+d=44\)
Xác định hệ số \(a,b,c,d\)
Vd: \(a=3;b=-2;c=\dfrac{2}{3};d=7\)
1. \(a=\dfrac{1}{3};b=1\)
2. \(a=\dfrac{1}{4};b=1;c=-1\)
3. \(b=1;c=\dfrac{1}{4};d=1\)
4. \(a=1;b=1;c=1;d=1\)
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Cho\(\frac{a}{b}\)=\(\frac{c}{d}\).CM\(\frac{4.a-3.b}{4.c-3.d}\)=\(\frac{4.a+3.b}{4.c+3.d}\)
b)\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a^2-c^2}{b^2-d^2}\)
a) Ta có: \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{4a}{4c}=\frac{3c}{3d}\)
Theo tín chất dãy tỉ số bằng nhau ta có:
\(\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a+3b}{4c+3d}=\frac{4a-3b}{4c-3d}\)(đpcm)
b) Ta có: \(\frac{a}{b}=\frac{c}{d}=>\frac{a^2}{b^2}=\frac{c^2}{d^2}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=>\frac{a^2+c^2}{b^2+d^2}=\frac{a^2-c^2}{b^2-d^2}\)(đpcm)
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