Bài 1: Cho (a+b-c)/c = (b+c-a)/a = (c+a-b)/b. Tính
P= (1+b/a) . (1+c/b) . (1+a/c)
Bài 2: Cho x/(y+z+t) = y/(z+t+x) = z/(t+x+y) = t/(x+t+z). Tính
(x+y)/(z+t) + (y+z)/(t+x) + (z+t)/(x+y) + (t+x)/(y+z)
Bài 1: áp dụng tính chất dãy tỉ số bằng nhau ta được:
(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(a+b+c)=(a+b+c)/(a+b+c0=1
Do đó: (a+b+c)/c=1 suy ra a+b+c=c suy ra a+b=c-c=0 nên a=b (1)
(b+c-a)/a=1 suy ra b+c-a=a suy ra a+c-a=a (b=a) suy ra c=a (2) Từ (1) và(2) ta có: a=b=c
Suy ra:P= (1+b/a).(1+c/b).(1+a/c)=(1+a/a).(1+a/a).(1+a/a)=(1+1).(1+1).(1+1)=2.2.2=8
Bài 2: bạn cũng áp dụng tính chất dãy tỉ bằng nhau rồi xét giống bài 1 là ra
a,Tìm x,y,z biết: \(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)
b,Cho \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh rằng: (\(\dfrac{a+b+c}{b+c+d}\))3=\(\dfrac{a}{d}\)
c,Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng: \(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)
d,Cho \(\dfrac{3x-2y}{4}\)=\(\dfrac{2z-4x}{3}\)=\(\dfrac{4y-3z}{2}\).Chứng minh rằng: \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{4}\)
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
a)cho a,b,c >0
CMR (a+1)(b+1)(a+c)(b+c)>=16abc
b)cho x,y,z>0 CMR x+y/z+y+z/x+z+x/y>= 6
c)cho a>=1, b>=1 CMR a căn b-1+b căn a-1 <=ab
Cho A = x-y / x+y
B = y-z/ y+z
C= z-x / z+x
chứng minh: (1+A). (1+B).(1+C) = (1-A).(1-B).(1-C)
cho x=(a-b)/(a+b); y=(b-c)/(b+c); z=(c-a)/(c+a)
cmr: (x+1)(y+1)(z+1)=(1-x)(1-y)(1-z)
Thay \(x=\frac{a-b}{a+b};y=\frac{b-c}{b+c};z=\frac{c-a}{c+a}\) vào (x + 1)(y + 1)(z + 1) và (1 - x)(1 - y)(1 - z) ta có:
\(\left(x+1\right)\left(y+1\right)\left(z+1\right)=\left(\frac{a-b}{a+b}+1\right)\left(\frac{b-c}{b+c}+1\right)\left(\frac{c-a}{c+a}+1\right)\)
\(=\frac{2a}{a+b}.\frac{2b}{b+c}.\frac{2c}{c+a}=\frac{2a.2b.2c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\left(1\right)\)
\(\left(1-x\right)\left(1-y\right)\left(1-z\right)=\left(1-\frac{a-b}{a+b}\right)\left(1-\frac{b-c}{b+c}\right)\left(1-\frac{c-a}{c+a}\right)\)
\(=\frac{2b}{a+b}.\frac{2c}{b+c}.\frac{2a}{c+a}=\frac{2b.2c.2a}{\left(a+b\right).\left(b+c\right).\left(c+a\right)}\left(2\right)\)
Từ (1) và (2) => đpcm
Cho x, y, z thỏa: x+y+z=a ; x^2+y^2+z^2=b ; 1/x+1/y+1/z=1/c Tính xy + yz +xz và x^3+y^3+z^3 theo a,b,c
ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)
\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)
Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)
cho x+y+z=a ;x^2+ y^2 + z^2=b^2 ; 1/x+1/y+1/z=1/c . Tinh x^3+y^3+z^3 theo a,b,c
cho a,b, c, x, y, z :{a/x+b/y+c/z=0;x/a+y/b+z/c=1
CMR:x^2/a^2+y^2/b^2+z^2/c^2=1
a) Cho x,y,z thuộc R. CMR: (x+y+z)²≤3(x²+y²+z²) b) cho a+b+c=1 và a,b,c≥-¼ Tìm giá trị lớn nhất của biểu thức A=√(4a+1) + √(4b+1) + √(4c+1)
a) (x + y + z)2 \(\le3\left(x^2+y^2+z^2\right)\)(1)
<=> \(x^2+y^2+z^2+2xy+2yz+2zx\le3x^2+3y^2+3z^2\)
<=> \(2x^2+2y^2+2z^2-2xy-2xz-2yz\ge0\)
<=> (x - y)2 + (y - z)2 + (z - x)2 \(\ge0\) (đúng)
=> (1) đúng "=" khi x = y = z
b) \(A=1\sqrt{4a+1}+1.\sqrt{4b+1}+1.\sqrt{4c+1}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4a+1+4b+1+4c+1\right)}\)
\(=\sqrt{3.\left[4\left(a+b+c\right)+3\right]}=\sqrt{21}\left(\text{vì }a+b+c=1\right)\)
"=" xảy ra <=> \(\dfrac{1}{\sqrt{4a+1}}=\dfrac{1}{\sqrt{4b+1}}=\dfrac{1}{\sqrt{4c+1}};a+b+c=1\)
<=> a = b = c = 1/3
cho a, b, c, x, y, z:{a/x+b/y+c/z=0;x/a+y/b+z/c=1
CMR:x^2/a^2+y^2/b^2+z^2/c^2=1
Ta có :\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
Lại có \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
=> \(\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)
=> \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xyc}{abc}+\frac{2ayz}{abc}+\frac{2bxz}{abc}=1\)
=> \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2}{abc}\left(xyc+ayz+bxz\right)=1\)
=> \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(\text{vì }xyc+ayz+bxz=0\right)\)(đpcm)
