Cho 2 < x < 3. Tìm GTNN của \(P=\dfrac{1}{\left(x-2\right)^2}+\dfrac{1}{\left(3-x\right)^2}+\dfrac{1}{\left(x-2\right)\left(3-x\right)}\)
cho x,y,z thỏa mãn \(x+y+z\le\dfrac{3}{2}\) . tìm GTNN của \(P=\dfrac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\dfrac{y\left(xz+1\right)^2}{y^2\left(xy+1\right)}+\dfrac{z\left(xy+1\right)^2}{x^2\left(yz+1\right)}\)
Áp dụng bất đẳng thức AM - GM:
\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).
Áp dụng bất đẳng thức AM - GM ta có:
\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).
Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).
Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)
\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).
Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)
Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)
\(\Rightarrow P\ge\dfrac{15}{2}\).
Vậy...
Áp dụng bất đẳng thức AM - GM:
P≥33√(xy+1)(yz+1)(zx+1)xyz.
Áp dụng bất đẳng thức AM - GM ta có:
xy+1=xy+14+14+14+14≥55√xy44.
Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.
Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412
⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.
Mà xyz≤(x+y+z)327=18
Nên (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258
⇒P≥152.
Tìm GTNN của các hàm số sau:
a) \(f\left(x\right)=5+x+\dfrac{1}{x}\left(x>4\right)\)
b) \(g\left(x\right)=\left(x+2\right)\left(3+\dfrac{1}{x}\right)\left(x>0\right)\)
c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2}{x+1}+2\right)^2\left(x\ne-1\right)\)
c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
Câu a muốn có min thì đề bài phải là \(x\ge4\) (có dấu "=")
Còn \(x>4\) thì chắc là đề sai
Tìm x.
\(1,\dfrac{3}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)=\dfrac{1}{4}\)
\(2,3\left(x-2\right)-4\left(x+2\right)=x+2\)
\(3,4x\left(x-1\right)+4x-2\left(x+1\right)=-2\)
\(4,x\left(x+2\right)-3\left(x-1\right)=3\left(x+1\right)\)
1. Tìm GTNN của \(y=x+\dfrac{1}{x}-5\) trên \(\left(0,+\infty\right)\)
2. Tìm GTNN của \(y=4x^2+\dfrac{1}{x}-4\) trên \(\left(0,+\infty\right)\)
3. Tìm GTLN của \(y=\dfrac{x^2+4}{x}\) trên \(\left(-\infty,0\right)\)
\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)
\(y_{min}=-3\) khi \(x=1\)
\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)
\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)
\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)
\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)
P=\(\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}\)
a) rút gọn P
b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
Tìm GTNN của mỗi biểu thức sau:
a) \(P=\left(x+30\right)^2+\left(y-4\right)^2+1975 \)
b)\(Q=\left(3x+1\right)^2+\left|2y-\dfrac{1}{3}\right|+\sqrt{5}\)
c)\(R=\dfrac{3}{1-x-x^2}\)
3 câu này bạn áp dụng cái này nhé.
`a^2 >=0 forall a`.
`|a| >=0 forall a`.
`1/a` xác định `<=> a ne 0`.
a: P=(x+30)^2+(y-4)^2+1975>=1975 với mọi x,y
Dấu = xảy ra khi x=-30 và y=4
b: Q=(3x+1)^2+|2y-1/3|+căn 5>=căn 5 với mọi x,y
Dấu = xảy ra khi x=-1/3 và y=1/6
c: -x^2-x+1=-(x^2+x-1)
=-(x^2+x+1/4-5/4)
=-(x+1/2)^2+5/4<=5/4
=>R>=3:5/4=12/5
Dấu = xảy ra khi x=-1/2
Tính
a)\(\left(\dfrac{\left(x-1\right)^2}{\left(3x+x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{x^2+x}{x^2+1}\)
b)\(\left(\dfrac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\dfrac{2x^2-x+10}{2\left(x^3+x^2+x+1\right)}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{2\left(x-1\right)}\right).\dfrac{2}{x-1}\)
c)\(\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right):\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
Tìm \(\left(x\right)\):
a) \(\left(x-\dfrac{1}{2}\right)^2\):\(\dfrac{2}{9}\) =\(\left(-1\dfrac{1}{3}\right)\):\(\left(\dfrac{1}{2}-x\right)\)
b) \(\dfrac{3.x-1}{4}\)=\(\dfrac{2.x-5}{3}\)
c) \(\dfrac{3}{-2}\)=\(\dfrac{x-3}{3.x+1}\)
Cú tui mấy ông bà ơi:)) Làm dc thì làm giúp tui nhen, cảm ơn nè
a) Bổ sung cho đầy đủ đề
b) (3x - 1)/4 = (2x - 5)/3
3(3x - 1) = 4(2x - 5)
9x - 3 = 8x - 20
9x - 8x = -20 + 3
x = -17
c) Điều kiện: x ≠ -1/3
3/(-2) = (x - 3)/(3x + 1)
3.(3x + 1) = -2(x - 3)
9x + 3 = -2x + 6
9x + 2x = 6 - 3
11x = 3
x = 3/11 (nhận)
Vậy x = 3/11