Cho \(\frac{a}{b}\)= \(\frac{c}{d}\), chứng minh \(\frac{5\cdot a+3\cdot b}{5\cdot a-3\cdot b}\)= \(\frac{5\cdot c+3\cdot d}{5\cdot c-3\cdot d}\)
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Tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). CMR
\(\frac{7\cdot a^3+3\cdot a\cdot b}{11\cdot a^2-8\cdot b^2}=\frac{7\cdot c^2+3\cdot c\cdot d}{11\cdot c^2+8\cdot d^2}\)
Đặt : \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\Rightarrow\frac{7b^2k^2+3bkb}{11b^2k^2-8b^2}=\frac{7d^2k^2+3dkd}{11d^2k^2-8d^2}\)
\(\Rightarrow\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}\)
\(\Rightarrow\frac{7k^2+3k}{11k^2-8}=\frac{7k^2+3k}{11k^2-8}\left(đpcm\right)\)
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\(\frac{2\cdot a+13\cdot b}{3\cdot a-7\cdot b}=\frac{2\cdot c+13\cdot d}{3\cdot c-7\cdot d}\)
CMR \(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)
\(\Leftrightarrow6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)
\(\Leftrightarrow-14ad+14bc=39ad-39bc\)
\(\Leftrightarrow-14\left(ad-bc\right)=39\left(ad-bc\right)\)
=>ad-bc=0
=>ad=bc
hay a/b=c/d
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Tìm a, b, c biết: \(\frac{3\cdot a-2\cdot b}{5}=\frac{2\cdot c-5\cdot a}{3}=\frac{5\cdot b-3\cdot c}{2}\)và a+b+c= -50
Tìm 3 số a,b,c biết: \(\frac{3\cdot a-2\cdot b}{5}=\frac{2\cdot c-5\cdot a}{3}=\frac{5\cdot b-3\cdot c}{2}\) và a+b+c=-50
a) Chứng minh biểu thức sau không phụ thuộc vào x:
\(\left(\frac{5\cdot a+b}{5\cdot a^2-a\cdot b}+\frac{5\cdot a-b}{5\cdot a^2-a\cdot b}\right)\div\frac{100\cdot a^2+4\cdot b^2}{25\cdot a^3-a\cdot b^2}\)
b) Tìm x; y sao cho \(x^3+y^3=3\cdot x\cdot y-1\)
a,frac{16}{15}cdotfrac{-5}{14}cdotfrac{54}{24}cdotfrac{56}{21}
b,5cdotfrac{7}{5} c,frac{1}{7}cdotfrac{5}{9}+frac{5}{9}cdotfrac{1}{7}+frac{5}{9}cdotfrac{3}{7}
d,4cdot11cdotfrac{3}{4}cdotfrac{9}{121}
e,frac{3}{4}cdotfrac{16}{9}-frac{7}{5}:frac{-21}{20}
g,2frac{1}{3}-frac{1}{3}cdotleft[frac{-3}{2}+left(frac{2}{3}+0,4cdot5right)right]
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\(a,\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(b,5\cdot\frac{7}{5}\) \(c,\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(d,4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(e,\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(g,2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
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) Ta có: 1615⋅−514⋅5424⋅56211615⋅−514⋅5424⋅5621
=1615⋅−514⋅94⋅83=1615⋅−514⋅94⋅83
=4⋅−13⋅47⋅3=4⋅−13⋅47⋅3
=12⋅−421=−4821=−167=12⋅−421=−4821=−167
b) Ta có: 5⋅75=355=75⋅75=355=7
c) Ta có: 17⋅59+59⋅17+59⋅3717⋅59+59⋅17+59⋅37
=59(17+17+37)=59(17+17+37)
=59⋅57=2563=59⋅57=2563
d) Ta có: 4⋅11⋅34⋅91214⋅11⋅34⋅9121
=4⋅11⋅3⋅94⋅121=2711=4⋅11⋅3⋅94⋅121=2711
e) Ta có: 34⋅169−75:−212034⋅169−75:−2120
=43+43=83=43+43=83
g) Ta có: 213−13⋅[−32+(23+0,4⋅5)]213−13⋅[−32+(23+0,4⋅5)]
=73−13⋅[−32+23+2]=73−13⋅[−32+23+2]
=73−13⋅76=73−13⋅76
=73−718=4218−718=3518
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Bài 22, Cho Afrac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot...cdotfrac{99}{100}
Bfrac{2}{3}cdotfrac{4}{5}cdotfrac{6}{7}cdot...cdotfrac{100}{101}
1/ So sánh A và B, A2 và A.B
2/ Chứng minh Afrac{1}{10}
Bài 21, Cho Afrac{1cdot3cdot5cdot...cdot4095}{2cdot4cdot6cdot...cdot4096}
Bfrac{2cdot4cdot6cdot...cdot4096}{1cdot3cdot5cdot...cdot4097}
1/ So sánh A2 và A.B
2/ Chứng minh Afrac{1}{64}
Bài 21, Cho Afrac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot...cdotfrac{2499...
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Bài 22, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)
\(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
1/ So sánh A và B, A2 và A.B
2/ Chứng minh A<\(\frac{1}{10}\)
Bài 21, Cho \(A=\frac{1\cdot3\cdot5\cdot...\cdot4095}{2\cdot4\cdot6\cdot...\cdot4096}\)
\(B=\frac{2\cdot4\cdot6\cdot...\cdot4096}{1\cdot3\cdot5\cdot...\cdot4097}\)
1/ So sánh A2 và A.B
2/ Chứng minh A<\(\frac{1}{64}\)
Bài 21, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{2499}{2500}\)Chứng minh A<\(\frac{1}{49}\)
Bài 22, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)
\(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{98}{99}\)
1/ So sánh A, B, C
2/Chứng minh \(A\cdot C< A^2< \frac{1}{10}\)
3/Chứng minh \(\frac{1}{15}< A< \frac{1}{10}\)
Bài 1: cho a,b,cge0 và a+b+c1. Chứng minh rằng :a,left(1-aright)cdotleft(1-bright)cdotleft(1-cright)ge8cdot acdot bcdot cb,16cdot acdot bcdot cge a+bc,frac{a}{1+a}+frac{2cdot b}{2+b}+frac{3cdot c}{3+c}lefrac{6}{7}Bài 2: cho a,b,c0 và a.b.c0 chứng minh rằng:frac{bcdot c}{a^2cdot b+a^2cdot c}+frac{acdot c}{b^2cdot c+b^2cdot a}+frac{acdot b}{c^2cdot a+c^2cdot b}gefrac{3}{2}
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Bài 1: cho \(a,b,c\ge0\) và a+b+c=1. Chứng minh rằng :
a,\(\left(1-a\right)\cdot\left(1-b\right)\cdot\left(1-c\right)\ge8\cdot a\cdot b\cdot c\)
b,\(16\cdot a\cdot b\cdot c\ge a+b\)
c,\(\frac{a}{1+a}+\frac{2\cdot b}{2+b}+\frac{3\cdot c}{3+c}\le\frac{6}{7}\)
Bài 2: cho a,b,c>0 và a.b.c=0 chứng minh rằng:
\(\frac{b\cdot c}{a^2\cdot b+a^2\cdot c}+\frac{a\cdot c}{b^2\cdot c+b^2\cdot a}+\frac{a\cdot b}{c^2\cdot a+c^2\cdot b}\ge\frac{3}{2}\)
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
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11 tháng 4 2021 lúc 21:02
Tính giá trị biểu thức sau :A acdotfrac{1}{2}+acdotfrac{1}{3}+acdotfrac{1}{4}với afrac{-4}{5}B frac{3}{4}cdot b+frac{4}{3}cdot b-frac{1}{2}cdot bvớibfrac{6}{19}C ccdotfrac{3}{4}+ccdotfrac{5}{6}-ccdotfrac{19}{12}với cfrac{2002}{2003}
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Tính giá trị biểu thức sau :
A = \(a\cdot\frac{1}{2}+a\cdot\frac{1}{3}+a\cdot\frac{1}{4}\)với \(a=\frac{-4}{5}\)
B = \(\frac{3}{4}\cdot b+\frac{4}{3}\cdot b-\frac{1}{2}\cdot b\)với\(b=\frac{6}{19}\)
C = \(c\cdot\frac{3}{4}+c\cdot\frac{5}{6}-c\cdot\frac{19}{12}\)với \(c=\frac{2002}{2003}\)
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0