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\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)
ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1
=> A ≥ 1
=> Min A =1 khi 1/3 ≤ x ≤ 2/3
Áp dụng bất đẳng thức cô si ta có :
\(x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3+\left(\sqrt{z}\right)^3\ge3\sqrt[3]{\left(\sqrt{xyz}\right)^3}=3\sqrt{xyz}\)Dấu "=" xảy ra khi :\(\sqrt{x}=\sqrt{y}=\sqrt{z}\)
Do đó :\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Vậy A=8
Ta có : \(3\sqrt{xyz}=\sqrt{x}^2+\sqrt{y}^3+\sqrt{z}^3\ge3\sqrt[3]{\sqrt{x}^3\sqrt{y}^3\sqrt{z}^3}=3\sqrt{x}\sqrt{y}\sqrt{z}=3\sqrt{xyz}.\)
Dấu = xảy ra
=> x =y =z
=> A = (1+1)(1+1)(1+1) =8
mk thấy nó sai sai . Tại sao 3\(\sqrt[3]{\sqrt{x}^3\sqrt{y}^3\sqrt{z}^3}\) = 3\(\sqrt{x}\sqrt{y}\sqrt{z}\)
Câu 1:
\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)
\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)
Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)
\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)
Và do $a,b\geq 3$ nên:
\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)
\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)
Cộng tất cả những BĐT trên ta có:
\(A\geq 2+14+62+2=80\) (đpcm)
Dấu "=" xảy ra khi $a=b=3$
Câu 2:
Bình phương 2 vế ta thu được:
\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)
\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)
\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)
\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)
Do đó pt đã cho vô nghiệm.
Dự đoán \(x=y=z=1\) ta tính được \(A=6+3\sqrt{2}\)
Ta sẽ c/m nó là GTLN của A
Thật vậy, ta cần chứng minh \(Σ\left(2+\sqrt{2}-2\sqrt{x}-\sqrt{1+x^2}\right)\ge0\)
\(\LeftrightarrowΣ\left(\frac{2\left(1-x\right)}{1+\sqrt{x}}+\frac{1-x^2}{\sqrt{2}+\sqrt{1+x^2}}\right)\ge0\)
\(\LeftrightarrowΣ\left(x-1\right)\left(1+\frac{1}{\sqrt{2}}-\frac{2}{1+\sqrt{x}}-\frac{x+1}{\sqrt{2}+\sqrt{1+x^2}}\right)+\left(1+\frac{1}{\sqrt{2}}\right)\left(3-x-y-z\right)\ge0\)
\(\LeftrightarrowΣ\left(x-1\right)^2\left(\frac{1}{\left(1+\sqrt{x}\right)^2}-\frac{x+1}{\sqrt{2}\left(\sqrt{2}+\sqrt{1+x^2}\right)\left(\sqrt{2}x+\sqrt{1+x^2}\right)}\right)+\left(1+\frac{1}{\sqrt{2}}\right)\left(3-x-y-z\right)\ge0\)
BĐT cuối đủ để chứng minh
\(\sqrt{2}\left(\sqrt{2}+\sqrt{1+x^2}\right)\left(\sqrt{2}x+\sqrt{1+x^2}\right)\ge\left(x+1\right)\left(1+\sqrt{x}\right)^2\)
Đặt \(1+x=2k\sqrt{x}\). Hence, theo Cauchy-Schwarz:
\(\sqrt{2}\left(\sqrt{2}+\sqrt{1+x^2}\right)\left(\sqrt{2}x+\sqrt{1+x^2}\right)\)
\(=\sqrt{2}\left(\sqrt{2}+\frac{1}{\sqrt{2}}\sqrt{2\left(1+x^2\right)}\right)\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\sqrt{2\left(1+x^2\right)}\right)\)
\(\ge\sqrt{2}\left(\sqrt{2}+\frac{x+1}{\sqrt{2}}\right)\left(\sqrt{2}x+\frac{x+1}{\sqrt{2}}\right)\)
\(=\frac{1}{\sqrt{2}}\left(x+3\right)\left(3x+1\right)=\frac{1}{\sqrt{2}}\left(3x^2+10x+3\right)\)
\(=\frac{1}{\sqrt{2}}\left(3\left(4k^2-2\right)x+10x\right)2\sqrt{2}x\left(3k^2+1\right)\)
Mặt khác \(\left(x+1\right)\left(1+\sqrt{x}\right)^2=\left(x+1\right)\left(x+1+2\sqrt{x}\right)\)
\(=2k\left(2k+2\right)x=4k\left(k+1\right)x\). Có nghĩa là ta cần phải c/m
\(3k^2+1\ge\sqrt{2}k\left(k+1\right)\Leftrightarrow\left(3-\sqrt{2}\right)k^2-2\sqrt{k}+1\ge0\)
Nó đúng theo AM-GM
\(\left(3-\sqrt{2}\right)k^2-\sqrt{2}k+1\ge\left(2\sqrt{3-\sqrt{2}}-\sqrt{2}\right)k\ge0\)
Hơi đẹp nhỉ nhưng xong r` đó :D
bunyakovsky:
\(\left(\sqrt{1+x^2}+\sqrt{2x}\right)^2\le2\left(x+1\right)^2\)
\(\Leftrightarrow\sqrt{1+x^2}+\sqrt{2}.\sqrt{x}\le\sqrt{2}\left(x+1\right)\)
tương tự :phần còn lại + thêm với\(\left(2-\sqrt{2}\right)\left(x+y+z\right)\)
Áp dụng bđt Cauchy-Schwarz:
\(A=\dfrac{1}{\sqrt{x\left(y+2z\right)}}+\dfrac{1}{\sqrt{y\left(z+2x\right)}}+\dfrac{1}{\sqrt{z\left(x+2y\right)}}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
\(=\dfrac{9}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
Áp dụng liên tiếp Bunyakovsky và AM-GM:
\(\left(\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left[x\left(y+2z\right)+y\left(z+2x\right)+z\left(x+2y\right)\right]\)
\(=3.3\left(xy+yz+xz\right)\)
Mà \(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=3\)
\(3.3\left(xy+yz+xz\right)\le3.3=9\)
\(\Leftrightarrow\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+z\sqrt{\left(x+2y\right)}\le\sqrt{9}=3\)
\(\Leftrightarrow A\ge\dfrac{9}{3}=3."="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
Bài 1:
a: \(A=\left|2a-1\right|-2a\)
TH1: a>=1/2
A=2a-1-2a=-1
TH2: a<1/2
A=1-2a-2a=1-4a
b: \(B=x-2y-\left|x-2y\right|\)
TH1: x>=2y
A=x-2y-x+2y=0
TH2: x<2y
A=x-2y+x-2y=2x-4y
c: \(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
\(A=x^2+x^2-4=2x^2-4\)
TH2: -2<x<2
\(A=x^2+4-x^2=4\)
d: \(D=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
TH1: x>5
\(D=2x-1-1=2x-2\)
TH2: x<5
D=2x-1+1=2x
a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\dfrac{1}{x^3\left(y+z\right)}+\dfrac{1}{y^3\left(x+z\right)}+\dfrac{1}{z^3\left(x+y\right)}\)
\(=\dfrac{x^2y^2z^2}{x^3\left(y+z\right)}+\dfrac{x^2y^2z^2}{y^3\left(x+z\right)}+\dfrac{x^2y^2z^2}{z^3\left(x+y\right)}\)
\(=\dfrac{y^2z^2}{x\left(y+z\right)}+\dfrac{x^2z^2}{y\left(x+z\right)}+\dfrac{x^2y^2}{z\left(x+y\right)}\)
\(\ge\dfrac{\left(xy+yz+xz\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{xy+yz+xz}{2}\ge\dfrac{3\sqrt[3]{\left(xyz\right)^2}}{2}=\dfrac{3}{2}=VP\)
Xảy ra khi \(x=y=z=1\)
b)Áp dụng BĐT Cauchy-Schwarz ta có:
\(y^2=\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\)
\(\le\left(1+1\right)\left(x-2+4-x\right)=4\)
\(\Rightarrow y^2\le4\Rightarrow y\le2\)
Khi \(x=3\)