câu 1 tìm A biết
\(A=\dfrac{a}{b+c}=\dfrac{c}{a+b}=\dfrac{b}{a+c}\)
câu 2
x∈Z để A∈Z
\(A=\dfrac{x+3}{x-2}\)
\(A=\dfrac{1-2x}{x+3}\)
nếu ai giải được mình cho 1 like
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30 tháng 12 2020
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
30 tháng 12 2020
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
TT
7 tháng 1 2021
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
23 tháng 12 2020
a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)
\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)
Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)
nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)
mà a+b+c=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)
Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)
b) Ta có: 2a=3b=5c
nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)
mà a+b-c=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)
Do đó:
\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)
Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)
9 tháng 12 2021
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
9 tháng 12 2021
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
KK
25 tháng 3 2017
Bải 3a
\(\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{-a}{2a}+\dfrac{b+c}{2a}+\dfrac{-b}{2b}+\dfrac{c+a}{2b}+\dfrac{-c}{2c}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow-\dfrac{3}{2}+\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge3\)
\(\Leftrightarrow\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge6\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{ca}{ac}}=2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge2+2+2=6\)
\(\Leftrightarrow\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
Bài 3b)
\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)
\(P=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\)( 1 )
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow\)\(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow P\ge\dfrac{3}{2}\)
Vậy \(P_{min}=\dfrac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c\)
25 tháng 3 2017
Bài 2:
a) \(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)
\(\Rightarrow\left(\dfrac{x-17}{33}-1\right)+\left(\dfrac{x-21}{29}-1\right)+\left(\dfrac{x}{25}-2\right)=0\)
\(\Rightarrow\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\right)=0\)
Mà \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\ne0\)
\(\Rightarrow x-50=0\)
\(\Rightarrow x=50\)
Vậy x = 50
21 tháng 1 2021
Bổ sung phần c và d luôn:
c, C = \(\dfrac{2}{5}\)
\(\Leftrightarrow\) \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{2}{5}\)
\(\Leftrightarrow\) 5(x2 - 1) = 2(2x2 + 3)
\(\Leftrightarrow\) 5x2 - 5 = 4x2 + 6
\(\Leftrightarrow\) x2 = 11
\(\Leftrightarrow\) x2 - 11 = 0
\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)
d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)
C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x2 + \(\dfrac{3}{2}\))
\(\Leftrightarrow\) 4(x2 + \(\dfrac{3}{2}\)) \(\in\) Ư(5)
Xét các TH:
4(x2 + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x2 = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)
4(x2 + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x2 = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)
4(x2 + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x2 = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)
4(x2 + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x2 = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)
Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z
Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)
1
Áp dụng tính chất dãy tỉ số bằng nhau
`=>a/(b+c)=c/(a+b)=b/(a+c)=(a+b+c)/(2a+2b+2c)=1/2`
`=>b+c=2a`
`=>a+b+c=3a`
Hoàn toàn tương tự:
`a+b+c=3b`
`a+b+c=3c`
`=>a=b=c`
`=>A=1/2+1/2+1/2=3/2`
2
`A in Z`
`=>x+3 vdots x-2`
`=>x-2+5 vdots x-2`
`=>5 vdots x-2`
`=>x-2 in Ư(5)={1,-1,5,-5}`
`+)x-2=1=>x=3(TM)`
`+)x-2=-1=>x=1(TM)`
`+)x-2=5=>x=7(TM)`
`+)x-2=-5=>x=-3(TM)`
Vậy với `x in {1,3,-3,7}` thì `A in Z`
`A in Z`
`=>1-2x vdots x+3`
`=>-2(x+3)+1+6 vdots x+3`
`=>7 vdots x+3`
`=>x+3 in Ư(7)={1,-1,7,-7}`
`+)x+3=1=>x=-2(TM)`
`+)x+3=-1=>x=-4(TM)`
`+)x+3=-7=>x=-10(TM)`
`+)x+3=7=>x=4(TM)`
Vậy `x in {2,-4,4,10}` thì `A in Z`