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\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
a, Ta có : \(f\left(x\right)-g\left(x\right)=h\left(x\right)\)hay
\(4x^2+3x+1-3x^2+2x-1=h\left(x\right)\)
\(\Rightarrow h\left(x\right)=x^2+5x\)
b, Đặt \(h\left(x\right)=x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy nghiệm của đa thức h(x) là x = -5 ; x = 0
Đặt \(k\left(x\right)=7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow7\left(x^2+2x+3x+6\right)=0\Leftrightarrow7\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)
Vậy nghiệm của đa thức k(x) là x = -3 ; x = -2
xin lỗi mọi người 1 tý nha cái phần c) ý ạ đề thì vậy như thế nhưng có cái ở phần biểu thức ở dưới ý là
\(\left(\frac{3^2}{6}-81\right)^3\) chuyển thành \(\left(\frac{3^3}{6}81\right)^3\)
bị sai mỗi thế thôi ạ mọi người giúp em với ạ
a, \(-\frac{187}{70}\)
b,\(\frac{27}{70}\)
c,\(\frac{53}{14}\)
d,\(\frac{27}{4}\)
e,1
f,\(\frac{23}{4}\)
g,-1
i,6
k,315
l,\(\frac{9}{2}\)
b) \(\left(x+\frac{1}{2}\right)^3:3=-\frac{1}{81}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{81}\right).3\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x+\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{6}\)
Vậy \(x=-\frac{5}{6}.\)
c) \(\frac{x-2}{2}=\frac{8}{x-2}\left(x\ne2\right).\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=8.2\)
\(\Rightarrow\left(x-2\right)^2=16\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm4\right)^2\)
\(\Rightarrow x-2=\pm4.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-2\right\}.\)
Chúc bạn học tốt!
Bài làm :
\(\frac{8}{9}-\left(\frac{-1}{3}\right)^2+\left(\frac{5}{6}\right)^{2020}\times\left(\frac{6}{5}\right)^{2020}\)
\(=\frac{8}{9}-\frac{1}{9}+\left(\frac{5}{6}\times\frac{6}{5}\right)^{2020}\)
\(=\frac{7}{9}+1\)
\(=\frac{16}{9}\)
Học tốt nhé
a) \(3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
b) \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^6\)
\(\Leftrightarrow x+1=6\Leftrightarrow x=5\)
c) \(\frac{81}{3x}=9\)
\(\Leftrightarrow3x=9\Leftrightarrow x=3\)
d) \(2^{x+1}+2^{x+2}=192\)
\(\Leftrightarrow2^x.2+2^x.4=192\)
\(\Leftrightarrow2^x.6=192\Leftrightarrow2^x=32\Leftrightarrow x=5\)
e) Ta có : \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}\Rightarrow\left(x-1\right)^{2020}+\left(y+2\right)^{2020}\ge0}\)
Mà \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Bài giải
a, \(3^{x+1}=243\)
\(3^{x+1}=3^5\)
\(\Rightarrow\text{ }x+1=5\)
\(\Rightarrow\text{ }x=4\)
b, \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\frac{1}{2^{x+1}}=\frac{1}{2^6}\)
\(2^{x+1}=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
c, \(\frac{81}{3x}=9\)
\(27x=81\)
\(x=3\)
d, \(2^{x+1}+2^{x+2}=192\)
\(2^{x+1}\left(1+2\right)=192\)
\(2^{x+1}\cdot3=192\)
\(2^{x+1}=64=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
e, \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}}\) với mọi x,y nên \(\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow\text{ }x=1\text{ ; }y=-2\)
Dài đấy :))
a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)
\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)
\(\Leftrightarrow\left|x-1\right|+8=9\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))
\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)
\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)
d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)
\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)
\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)
\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)
Vậy ta xét hai trường hợp sau :
1. \(x\ge-\frac{3}{16}\)
(*) <=>\(7x-2=4x+\frac{3}{4}\)
\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)
\(\Leftrightarrow3x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)
2. \(x< -\frac{3}{16}\)
(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)
\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)
\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)
\(\Leftrightarrow11x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)
Vậy x = 11/12
e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)
\(\Leftrightarrow x+1=4040\)
\(\Leftrightarrow x=4039\)
F = | 2x - 2 | + | 2x - 2003 |
F = | 2x - 2 | + | -( 2x - 2003 ) |
F = | 2x - 2 | + | 2003 - 2x |
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
F = | 2x - 2 | + | 2003 - 2x | ≥ | 2x - 2 + 2003 - 2x | = | 2001 | = 2001
Đẳng thức xảy ra khi ab ≥ 0
=> ( 2x - 2 )( 2003 - 2x ) ≥ 0
Xét hai trường hợp :
1/ \(\hept{\begin{cases}2x-2\ge0\\2003-2x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}2x\ge2\\-2x\ge-2003\end{cases}}\Rightarrow\hept{\begin{cases}x\ge1\\x\le\frac{2003}{2}\end{cases}\Rightarrow}1\le x\le\frac{2003}{2}\)
2/ \(\hept{\begin{cases}2x-2\le0\\2003-2x\le0\end{cases}}\Rightarrow\hept{\begin{cases}2x\le2\\-2x\le-2003\end{cases}}\Rightarrow\hept{\begin{cases}x\le1\\x\ge\frac{2003}{2}\end{cases}}\)( loại )
Vậy MinF = 2001 <=> \(1\le x\le\frac{2003}{2}\)
G = | 2x - 3 | + 1/2| 4x - 1 |
G = | 2x - 3 | + | 2x - 1/2 |
G = | -( 2x - 3 ) | + | 2x - 1/2 |
G = | 3 - 2x | + | 2x - 1/2 |
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
G = | 3 - 2x | + | 2x - 1/2 | ≥ | 3 - 2x + 2x - 1/2 | = | 5/2 | = 5/2
Đẳng thức xảy ra khi ab ≥ 0
=> ( 3 - 2x )( 2x - 1/2 ) ≥ 0
Xét 2 trường hợp :
1/ \(\hept{\begin{cases}3-2x\ge0\\2x-\frac{1}{2}\ge0\end{cases}}\Rightarrow\hept{\begin{cases}-2x\ge-3\\2x\ge\frac{1}{2}\end{cases}\Rightarrow}\hept{\begin{cases}x\le\frac{3}{2}\\x\ge\frac{1}{4}\end{cases}}\Rightarrow\frac{1}{4}\le x\le\frac{3}{2}\)
2/ \(\hept{\begin{cases}3-2x\le0\\2x-\frac{1}{2}\le0\end{cases}}\Rightarrow\hept{\begin{cases}-2x\le-3\\2x\le\frac{1}{2}\end{cases}\Rightarrow}\hept{\begin{cases}x\ge\frac{3}{2}\\x\le\frac{1}{4}\end{cases}}\)( loại )
=> MinG = 5/2 <=> \(\frac{1}{4}\le x\le\frac{3}{2}\)
H = | x - 2018 | + | x - 2019 | + | x - 2020 |
H = | x - 2019 | + [ | x - 2018 | + | x - 2020 | ]
H = | x - 2019 | + [ x - 2018 | + | -( x - 2020 ) | ]
H = | x - 2019 | + [ | x - 2018 | + | 2020 - x | ]
Ta có : | x - 2019 | ≥ 0 ∀ x
| x - 2018 | + | 2020 - x | ≥ | x - 2018 + 2020 - x | = | 2 | = 2 ( BĐT | a | + | b | ≥ | a + b | )
=> | x - 2019 | + [ | x - 2018 | + | 2020 - x | ] ≥ 2
Đẳng thức xảy ra <=> \(\hept{\begin{cases}\left|x-2019\right|=0\\\left(x-2018\right)\left(2020-x\right)\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2019\\2018\le x\le2020\end{cases}}\)
=> x = 2019
=> MinH = 2 <=> x = 2019
đáp án
2011,9