bai2:tim x biet
x=\(\sqrt{3}\)
x=\(\sqrt{8}\)
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ND
1
TN
10 tháng 5 2016
\(x-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x-x}{2007}=\frac{2006}{2007}\)
\(\frac{2006x}{2007}=\frac{2006}{2007}\Rightarrow2006x=2006\)
=>x=1
PT
0
K
5
2 tháng 6 2017
\(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x+3}}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-8+5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
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
2 tháng 6 2017
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
TD
11 tháng 2 2018
a)\(8\sqrt{x}-3\sqrt{\frac{4}{81}}=5,2\)
\(\Rightarrow8\sqrt{x}-3.\frac{2}{9}=5,2\)
\(\Rightarrow8\sqrt{x}-\frac{2}{3}=5,2\)
\(\Rightarrow8\sqrt{x}=5,2+\frac{2}{3}\)
\(\Rightarrow8\sqrt{x}=\frac{40}{3}\)
\(\Rightarrow\sqrt{x}=\frac{40}{3}:8\)
\(\Rightarrow\sqrt{x}=\frac{5}{3}\)
\(\Rightarrow x=\frac{25}{9}\)
b)\(12-3x^2=10+\sqrt{\frac{25}{16}}\)
\(\Rightarrow12-3x^2=10+\frac{5}{4}\)
\(\Rightarrow12-3x^2=11,25\)
\(\Rightarrow3x^2=12-11,25\)
\(\Rightarrow3x^2=0,75\)
\(\Rightarrow x^2=0,25\)
\(\Rightarrow x=\sqrt{0,25}\)
\(\Rightarrow x=0,5\)
ND
1 tháng 8 2018
\(\dfrac{2\sqrt{X}-9}{x-5\sqrt{X}+6}-\dfrac{\sqrt{X}+3}{\sqrt{X}-2}-\dfrac{2\sqrt{X}+1}{3-\sqrt{X}}\) \(\left(X\ne2;X\ne3,X\ge0\right)\)
\(=\dfrac{2\sqrt{X}-9-\left(\sqrt{X}+3\right)\left(\sqrt{X}-3\right)+\left(2\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)
\(=\dfrac{2\sqrt{X}-9-X+9+2X-4\sqrt{X}+\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)
\(=\dfrac{X-\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}=\dfrac{X-2\sqrt{X}+\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)
\(=\dfrac{\sqrt{X}\left(\sqrt{X}-2\right)+\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}=\dfrac{\left(\sqrt{X}-2\right)\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}=\dfrac{\sqrt{X}+1}{\sqrt{X}-3}\)
\(C=\dfrac{\sqrt{X}+1}{\sqrt{X}-3}< 1\)
\(\Rightarrow\dfrac{\sqrt{X}+1-\sqrt{X}+3}{\sqrt{X}-3}< 0\)
\(\Rightarrow\dfrac{4}{\sqrt{X}+3}< 0\) ( VÔ LÍ)
⇒ Không có X thỏa mãn
1 tháng 8 2018
sai dấu tỷ ới , hình như là \(\dfrac{4}{\sqrt{X}-3}< 0\) mà
x=\(\sqrt{3}\approx1,7320508\)
x=\(\sqrt{8}\approx2,828427125\)
chẳng đời nào lại có cái bài như thế này như thế ko cần tìm x nữa đâu
Có mà
\(\sqrt{x}=3\)
tìm x