Cho \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(A=\dfrac{x+y-z}{x-y+z};B=\dfrac{2x+3y+z}{x-2y-3z}\)
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
do x,y,z≥0 nên x2≥0 , y+z≥0
áp dụng bất đẳng thức cosi cho 2 số dương \(\dfrac{x^2}{y+z}\) và y+z/4
x^2/y+z +(y+z)/4≥2\(\sqrt{\dfrac{x^2}{y+z}.\dfrac{\left(y+z\right)}{4}}\) =x (1)
y^2/x+z+(x+z)/4≥2\(\sqrt{\dfrac{y^2}{x+z}.\dfrac{x+z}{4}}\) =y (2)
z^2/y+x+(y+x)/4≥2\(\sqrt{\dfrac{z^2}{y+x}.\dfrac{y+x}{4}}\) =z (3)
từ (1)(2)(3)
➜\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)+(y+z/4)+(z+x)/4+(x+y)/4 ≥ x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) +(a+b+c)/2 ≥x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥ (x+y+z)/2
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥1 (vì x+y+z=2)
vậy giá trị nhỏ nhất của \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) =1
Nham ko phai Nesbit, Cauchy-Schwarz ra luon
Tìm x,y,z biết:
a, x : y : z = 10 : 3 : 4 và x + 2y - 3z = -20
b, \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) và \(\dfrac{y}{5}\) = \(\dfrac{z}{4}\) và x - y + z = -49
c, \(\dfrac{x}{2}\)= \(\dfrac{y}{3}\) =\(\dfrac{z}{4}\) và xy + \(z^2\)= 88
d, \(\dfrac{x}{5}\)= \(\dfrac{y}{7}\) = \(\dfrac{z}{3}\) và \(x^2\) + \(y^2\) + \(z^2\) = 415
Giải hộ mk nha
1) Rút gọn bt:
(x+y+z)3+(x-y-z)3+(y-x-z)3+(z-y-x)3
2)Tìm x,y,z t/m: 9x2+y2+2z2-18x+4z-6y+20=0
3)Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\)=1 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\)=0 . CMR:
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)=1
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}\)và 2x+y-z=81
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}\)và 5x-y+3z=124
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và x.y.z=810
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}\)và\(x^2.y^2.z^2=288^2\)
a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
d.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)
Thế vào \(x^2y^2z^2=288^2\)
\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)
\(\Rightarrow\left(k^2\right)^3=64\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)
a) Chứng minh rằng nếu 2(x+y) = 5(y+z) = 3(z+x)
Thì \(\dfrac{x-y}{4}=\dfrac{y-z}{5}\)
b) Cho \(x^2=yz\) . Chứng minh rằng \(\dfrac{x^2+y^2}{y^2+z^2}=\dfrac{x}{z}\)
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
x12=y9=z5=k" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x12=y9=z5=k
x5=y7=z3=x225=y249=z29" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Bài 4:
a) \(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{3}và\) x-y+z=18
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và x+2y-3z=-20
a) Áp dụng tính chất của dãy tỉ số bằng nhau:
`x/2=y/6=z/3=(x-y+z)/(2-6+3)=18/(-1)=-18`
`=>x=-36`
`y=-108`
`z=-54`
b) Áp dụng tính chất của dãy tỉ số bằng nhau:
`x/2=y/3=z/4=(x+2y-3z)/(2+2.3-3.4)=(-20)/(-4)=5`
`=>x=10`
`y=15`
`z=20`.
\(a.\)
\(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{x-y+z}{2-6+3}=\dfrac{18}{-1}=-18\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot\left(-18\right)=-36\\y=6\cdot\left(-18\right)=-108\\z=3\cdot\left(-18\right)=-54\end{matrix}\right.\)
\(b.\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{20}{-4}=-5\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot\left(-5\right)=-10\\y=3\cdot\left(-5\right)=-5\\z=4\cdot\left(-5\right)=-20\end{matrix}\right.\)
a) Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{x-y+z}{2-6+3}=\dfrac{18}{-1}=-18\)
Do đó:
x=-36; y=-108; z=-54
Tìm x,y,z biết:
a) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và x-y+z=-21
b)\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(x^2-2y^2+z^2=44\)
\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)
⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)
⇒x=70;y=105;z=84
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)⇒\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}=\dfrac{x^2-2y^2+z^2}{4-18+25}=\dfrac{44}{11}=4\)
⇒x=8;y=12;z=20