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NT
1
9 tháng 6 2022
a: \(3x^2-5x+7\)
\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{7}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}+\dfrac{59}{36}\right)\)
\(=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{59}{12}\ge\dfrac{59}{12}\)
Dấu '=' xảy ra khi x=5/6
c: \(\left(x-3\right)^2+\left(x-2\right)^2\)
\(=x^2-6x+9+x^2-4x+4\)
\(=2x^2-10x+13\)
\(=2\left(x^2-5x+\dfrac{13}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{1}{4}\right)\)
\(=2\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=5/2
d: \(\left(x-1\right)\left(x+3\right)+11\)
\(=x^2+2x-3+11\)
\(=x^2+2x+8=\left(x+1\right)^2+7\ge7\)
Dấu '=' xảy ra khi x=-1
17 tháng 7 2022
1: \(=75\left(27+25-2\right)=75\cdot50=3750\)
2: \(=15\left(23+37\right)+55=15\cdot60+55=955\)
3: \(=36\cdot14+36\cdot17+36\cdot69\)
\(=36\cdot100=3600\)
4: \(=200\cdot\left(32+68\right)=200\cdot100=20000\)
TV
2
26 tháng 2 2020
a) 11+(15-x)=1
⇔15-x=-10
hay x=25
Vậy: x=25
b) 2x-35=15
⇔2x=50
hay x=25
Vậy: x=25
c) 2|x+5|=12
⇔|x+5|=6
\(\Leftrightarrow\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
Vậy: x∈{1;-11}
8 tháng 4 2020
A/ \( 11+(15-x)=1\)\(\Leftrightarrow11+15-x=1\Leftrightarrow x=11+15-1=25\)
KL: ...........
B/ \(2x-35=15\Leftrightarrow2x=15+35=50\Leftrightarrow x=25\)
KL: ............
C/ \(2\left|x+5\right|=12\Leftrightarrow\left|x+5\right|=6\Leftrightarrow\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
KL: .............
N
3 tháng 3 2019
1.ĐK: \(x\ge\dfrac{1}{4}\)
bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)
\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)
\(\Leftrightarrow20x^2-x-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)
2.ĐK: \(-2\le x\le\dfrac{5}{2}\)
bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)
\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)
\(\Leftrightarrow x^2< -x^2+x+6\)
\(\Leftrightarrow-2x^2+x+6>0\)
\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)
3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)
.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)
\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)
*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)
*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)
CW
24 tháng 6 2016
a/ \(\frac{15}{x}-\frac{1}{3}=\frac{28}{51}\)
\(\frac{15}{x}=\frac{28}{51}+\frac{1}{3}\)
\(\frac{15}{x}=\frac{15}{17}\)
\(x=15:\frac{15}{17}\)
\(x=17\)
b) \(\frac{x}{20}-\frac{2}{5}=10\)
\(\frac{x}{20}=10+\frac{2}{5}\)
\(\frac{x}{20}=\frac{52}{5}\)
\(x=\frac{52}{5}\cdot20\)
\(x=208\)
c) \(x+\frac{18}{23}=2\frac{1}{3}\)
\(x+\frac{18}{23}=\frac{7}{3}\)
\(x=\frac{7}{3}-\frac{18}{23}\)
\(x=\frac{107}{69}\)
d) \(\frac{7}{11}< x-\frac{1}{7}< \frac{10}{13}\)
\(\Rightarrow\frac{7}{11}+\frac{1}{7}< x< \frac{10}{13}\)
\(\frac{60}{77}< x< \frac{60}{78}\)
Đến đây .....bí!
e) Tớ bỏ luôn đc ko.
TQ
24 tháng 6 2016
D) 7/11<X-1/7<10/13
<=> 7/11+1/7<x< 10/13+1/7
<=> 60/77< x< 83/91
<=> 5460/1001 <x< 6391/1001
vậy X thuộc tập hợp các phÂN số lớn hơn 5460/1001 và bé hơn 913/1001
vd : Y/1001 trong đó y là 5461;5462;5463...6389;6390
KK
1
31 tháng 5 2022
1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)
2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
=>x+1=0
hay x=-1