\(\frac{1}{30-x}+\frac{1}{x-4}+\frac{11}{\sqrt{\left(30-x\right)\left(x-4\right)}}\)
Tim gia tri nho nhat
tim gia tri nho nhat cua
\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-\left(1+x^2y^2\right)^2\)
\(A=\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}+\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-x^4y^4-2x^2y^2-1\)
Áp dụng Côsi
\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}.2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\)
\(\frac{1}{4}\left(x^{16}+y^{16}+1+1+1+1+1+1\right)\ge\frac{1}{4}.8\sqrt[8]{x^{16}y^{16}}=2x^2y^2\)
\(\Rightarrow A+\frac{6}{4}\ge x^4y^4+2x^2y^2-x^4y^4-2x^2y^2-1=-1\)
\(\Rightarrow A\ge-1-\frac{6}{4}=-\frac{5}{2}\)
Dấu "=" xảy ra khi và chỉ khi \(x^2=y^2=1\)
Vậy GTNN của A là -2,5 khi x2 = y2 = 1
Cho x,y la cac so thuc duong. Tim gia tri nho nhat cua bieu thuc:
\(P=\frac{xy}{x^2+y^2}+\left(\frac{1}{x}+\frac{1}{y}\right)\sqrt{2\left(x^2+y^2\right)}\)
Hình như đề sai rùi bạn ơi !
Phải sửa xy/x^2+y^2 thành x^2+y^2/xy hoặc cái gì khác
Vì xy/x^2+y^2 chỉ có GTLN chứ ko có GTNN đâu
Mk nói có gì sai thì thông cảm nha !
đề không sai đâu bạn à. Đây là đề toán chuyên ở tỉnh mình mà
Theo B.C.S ta có \(\sqrt{2\left(x^2+y^2\right)}\)\(\ge\)(\(\sqrt{\left(x+y\right)^2}\)\(=x+y\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)\sqrt{2\left(x^2+y^2\right)}\ge\left(\frac{1}{x}+\frac{1}{y}\right)\left(x+y\right)=2+\frac{x^2+y^2}{xy}\)
\(\Leftrightarrow\)\(P\ge2+\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}+\frac{3\left(x^2+y^2\right)}{4xy}\)
\(\Leftrightarrow\)\(P\ge2+2\sqrt{\frac{xy}{x^2+y^2}\times\frac{x^2+y^2}{4xy}}\)\(+\frac{3\times2xy}{4xy}\)
\(\Leftrightarrow\)\(P\ge2+1+\frac{3}{2}=\frac{9}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow\)x=y
cho x;y;z duong thoa man xyz=1
tim gia tri nho nhat cua \(\frac{1}{x^2\left(y+z\right)}+\frac{1}{y^2\left(x+z\right)}+\frac{1}{z^2\left(x+y\right)}\)
Cho A \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}+1}{X+\sqrt{X}}\right):\frac{\sqrt{X}+1}{X}\)
a)Tìm dk cua x de A co nghia
b)Rut gon A
c)Tim gia tri nho nhat cua A
cho bieu thuc \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x+2}}{x-2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
a) rut gon P
b) tim gia tri lon nhat cua P
\(ĐKXĐ:0\le x\ne x\)
a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)
\(P=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(P=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}.\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(\Rightarrow MAX_P=\frac{1}{4}\text{ khi }x=\frac{1}{4}\)
rut gon M
\(M=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right).\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
xac dinh x de m dat gia tri nho nhat
\(M=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right).\)\(\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\frac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\sqrt{x}^3-1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)\(.\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\frac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}-1}\right).\)\(\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(2x+\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(2x+\sqrt{x}-1\right)}\)\(-\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-1\right)}\)\(+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)\(+\frac{\sqrt{x}}{2\sqrt{x}+1}\)
\(=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}}{2\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}+1}\)
Khổ nỗi đến đây tắc, cậu nghĩ ra lối thoát chưa hay tớ sai chỗ nào nhỉ ?
1 cho 3 so thuc duong thoa man x^2010+y^2010+z^2010=3 tim gia tri lon nhat cua x^2+y^2+z^2
2 cho a;b;c duong c/m \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>hoac=3\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)
3 tim gia tri nho nhat cua \(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ac+a^2}\) voi a+b+c=1
4 cho a;b;c;d va A;B;C;D la cac so duong thoa man \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)C/ M \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)
5 tim gia tri lon nhat cua \(\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
6 phan tich da thuc thanh nhan tu \(y-5x\sqrt{y}+6x^2\)
7 cho x;y;z>0 xy+yz+xz=1 tinh \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
8 cho a;b;c >0 c/m \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}
pn oi nhieu the nay ai ma giai cho het dc
bài lớp mấy mà nhìn ghê quá zật bạn..................Nhìu quá
tim gia tri nho nhat cua A= \(\frac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}\)
P=\(\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\\ \)
a, rut gon
b, tim x de P<\(\frac{15}{4}\)
c, tim gia tri lon nhat cua P
\(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x+3}}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-8+5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
b)Để \(P< \frac{15}{4}\)thì \(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)
Ta có:\(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)
\(\Leftrightarrow\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}-\frac{15}{4}< 0\)
\(\Leftrightarrow\frac{12\sqrt{x}+32-15\sqrt{x}-30}{4\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\frac{-\left(3\sqrt{x}+2\right)}{4\sqrt{x}+8}< 0\)
Vì \(x\ge0;x\ne1\)
Do đó \(0< 4\sqrt{x}+8\)
Mà \(-\left(3\sqrt{x}+2\right)< 0\)
Vậy \(P< \frac{15}{4}\left(đpcm\right)\)
c)Ta có:\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{3\sqrt{x}+6+2}{\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}+\frac{2}{2\sqrt{x}+2}\)
\(\Leftrightarrow P=3+\frac{2}{\sqrt{x}+2}\)
Vì \(x\ge0;x\ne1\Rightarrow\frac{2}{\sqrt{x}+2}\le1\)
Do đó \(P\le4\Leftrightarrow x=1\)
Vậy Max P=4 khi x=1
P=3x+3√x−9(√x−1)(√x+2) +√x+3√x+2 −√x−2√x−1
P=3x+3√x−9(√x−1)(√x+2) +(√x+3)(√x−1)(√x+2)(√x−1) −x−4(√x−1)(√x+2)
P=3x+3√x−9+x+2√x−3−x+4(√x−1)(√x+2)
P=3x−8+5√x(√x−1)(√x+2)
P=3x−3√x+8√x−8(√x−1)(√x+2)
P=(3√x+8)(√x−1)(√x−1)(√x+2)
P=(3√x+8)(√x+2)
b)Để P<154 thì (3√x+8)(√x+2) <154
Ta có:(3√x+8)(√x+2) <154
⇔(3√x+8)(√x+2) −154 <0
⇔12√x+32−15√x−304(√x+2) <0
⇔−(3√x+2)4√x+8 <0
Vì x≥0;x≠1
Do đó 0<4√x+8
Mà −(3√x+2)<0
Vậy P<154 (đpcm)
c)Ta có:P=(3√x+8)(√x+2)
⇔P=3√x+6+2(√x+2)
⇔P=3(√x+2)(√x+2) +22√x+2
⇔P=3+2√x+2
Vì x≥0;x≠1⇒2√x+2 ≤1
Do đó