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Bài 3:
a: \(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\cdots\left|x+\frac{1}{2019\cdot2020}\right|=2020x\) (1)
=>2020x>=0
=>x>=0
Phương trình (1) sẽ trở thành:
\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{2019\cdot2020}=2020x\)
=>\(2020x=2019x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\right)\)
=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\)
=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2019}-\frac{1}{2020}\)
=>\(x=1-\frac{1}{2020}=\frac{2019}{2020}\)
b: \(\left|x+\frac{1}{1\cdot3}\right|+\left|x+\frac{1}{3\cdot5}\right|+\cdots+\left|x+\frac{1}{197\cdot199}\right|=100x\) (2)
=>100x>=0
=>x>=0
(2) sẽ trở thành: \(x+\frac{1}{1\cdot3}+x+\frac{1}{3\cdot5}+\cdots+x+\frac{1}{197\cdot199}=100x\)
=>\(100x=99x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{197\cdot199}\right)\)
=>\(x=\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{197}-\frac{1}{199}\right)=\frac12\left(1-\frac{1}{199}\right)\)
=>\(x=\frac12\cdot\frac{198}{199}=\frac{99}{199}\)
c: \(\left|x+\frac12\right|+\left|x+\frac16\right|+\left|x+\frac{1}{12}\right|+\cdots+\left|x+\frac{1}{110}\right|=11x\left(3\right)\)
=>11x>=0
=>x>=0
(3) sẽ trở thành:
\(11x=x+\frac12+x+\frac16+\ldots+x+\frac{1}{110}\)
=>\(11x=10x+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)
=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)
=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\) (nhận)
Bài 2:
a: \(\left|5-\frac23x\right|\ge0\forall x;\left|\frac23y-4\right|\ge0\forall y\)
Do đó: \(\left|5-\frac23x\right|+\left|\frac23y-4\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}5-\frac23x=0\\ \frac23y-4=0\end{cases}\Rightarrow\begin{cases}\frac23x=5\\ \frac23y=4\end{cases}\Rightarrow\begin{cases}x=5:\frac23=\frac{15}{2}\\ y=4:\frac23=6\end{cases}\)
b: \(\left|\frac23-\frac12+\frac34x\right|=\left|\frac34x+\frac16\right|\ge0\forall x\)
\(\left|1,5-\frac34-\frac32y\right|=\left|\frac34-\frac32y\right|\ge0\forall y\)
Do đó: \(\left|\frac34x+\frac16\right|+\left|\frac34-\frac32y\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}\frac34x+\frac16=0\\ \frac34-\frac32y=0\end{cases}\Rightarrow\begin{cases}\frac34x=-\frac16\\ \frac32y=\frac34\end{cases}\Rightarrow\begin{cases}x=-\frac16:\frac34=-\frac16\cdot\frac43=-\frac{4}{18}=-\frac29\\ y=\frac34:\frac32=\frac24=\frac12\end{cases}\)
c: \(\left|x-2020\right|\ge0\forall x;\left|y-2021\right|\ge0\forall y\)
Do đó: \(\left|x-2020\right|+\left|y-2021\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}x-2020=0\\ y-2021=0\end{cases}\Rightarrow\begin{cases}x=2020\\ y=2021\end{cases}\)
d: \(\left|x-y\right|\ge0\forall x,y\)
\(\left|y+\frac{21}{10}\right|\ge0\forall y\)
Do đó: \(\left|x-y\right|+\left|y+\frac{21}{10}\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ y+\frac{21}{10}=0\end{cases}\Rightarrow x=y=-\frac{21}{10}\)
Bài 1:
a: \(\left|\frac32x+\frac12\right|=\left|4x-1\right|\)
=>\(\left[\begin{array}{l}4x-1=\frac32x+\frac12\\ 4x-1=-\frac32x-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}4x-\frac32x=\frac12+1\\ 4x+\frac32x=-\frac12+1\end{array}\right.\)
=>\(\left[\begin{array}{l}\frac52x=\frac32\\ \frac{11}{2}x=\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32:\frac52=\frac35\\ x=\frac12:\frac{11}{2}=\frac{1}{11}\end{array}\right.\)
b: \(\left|\frac75x+\frac12\right|=\left|\frac43x-\frac14\right|\)
=>\(\left[\begin{array}{l}\frac75x+\frac12=\frac43x-\frac14\\ \frac75x+\frac12=\frac14-\frac43x\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac75x-\frac43x=-\frac14-\frac12\\ \frac75x+\frac43x=\frac14-\frac12\end{array}\right.\)
=>\(\left[\begin{array}{l}\frac{1}{15}x=-\frac34\\ \frac{41}{15}x=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac34:\frac{1}{15}=-\frac34\cdot15=-\frac{45}{4}\\ x=-\frac14:\frac{41}{15}=-\frac14\cdot\frac{15}{41}=-\frac{15}{164}\end{array}\right.\)
c: \(\left|\frac54x-\frac72\right|-\left|\frac58x+\frac35\right|=0\)
=>\(\left|\frac54x-\frac72\right|=\left|\frac58x+\frac35\right|\)
=>\(\left[\begin{array}{l}\frac54x-\frac72=\frac58x+\frac35\\ \frac54x-\frac72=-\frac58x-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x-\frac58x=\frac35+\frac72\\ \frac54x+\frac58x=-\frac35+\frac72\end{array}\right.\)
=>\(\left[\begin{array}{l}\frac58x=\frac{41}{10}\\ \frac{15}{8}x=\frac{29}{10}\e...
Ta có \(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{x+z}\)
=> \(\frac{xyz}{xz+yz}=\frac{xyz}{xy+xz}=\frac{xyz}{xy+yz}\)
=> \(xz+yz=xy+xz=xy+yz\)(vì x ; y ;z \(\ne0\Leftrightarrow xyz\ne0\))
=> \(\hept{\begin{cases}xz+yz=xy+xz\\xy+xz=xy+yz\\xz+yz=xy+yz\end{cases}}\Rightarrow\hept{\begin{cases}yz=xy\\xz=yz\\xz=xy\end{cases}}\Rightarrow\hept{\begin{cases}z=x\\x=y\\y=z\end{cases}}\Rightarrow x=y=z\)
Khi đó M = \(\frac{x^2+y^2+z^2}{xy+yz+zx}=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1\left(\text{vì }x=y=z\right)\)
a)\(A=12-\left|x-3\right|-\left|y+7\right|\)
\(-\left|x-3\right|\le0;-\left|y+7\right|\le0\)
\(\Rightarrow A\le12-0-0=12\)
Vậy Max A = 12 <=> x = 3 ; y = -7
b)\(B=-\left(x-2018\right)^6-1\)
\(-\left(x-2018\right)^6\le0\)
\(B\le0-1=-1\)
Vậy Max B = -1 <=> x = 2018
a) \(A=12-\left|x-3\right|-\left|y+7\right|\)
Nhận thấy: \(\left|x-3\right|\ge0;\)\(\left|y+7\right|\ge0\)
suy ra: \(A=12-\left|x-3\right|-\left|y+7\right|\le12\)
Vậy MIN A = 12
Dấu "=" xảy ra <=> \(x=3;y=-7\)
b) \(B=-\left(x-2018\right)^6-1\)
Nhận thấy: \(\left(x-2018\right)^6\ge0\)
suy ra: \(B=-\left(x-2018\right)^2-1\le-1\)
Vậy MIN B = -1
Dấu "=" xảy ra <=> \(x=2018\)
c) \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\)
Nhận thấy: \(\left|x+8\right|\ge0\) \(\left(3y+7\right)^{2016}\ge0\)
suy ra: \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\le\frac{20}{7}\)
Vậy MIN C = 20/7
Dấu "=" xảy ra <=> \(x=-8;y=-\frac{7}{3}\)
a) Ta có 2011 = x => 2012 = x + 1
Thay x + 1 = 2012 vào biểu thức ta dc:
x5 - (x + 1)x4 + (x + 1)x3 - (x+1)x2 + (x+1)x - 2012
= x5 - x5 - x4 + x4 + x3 - x3 - x2 + x2 + x - 2012 = x - 2012 = 2011 - 2012 = -1
Vậy giá trị của biểu thức là -1 khi x = 2011
b) Ta có : (x - 1)60 + (y + 2)90 = 0 <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Thay x = 1 và y = -2 vào biểu thức ta dc: 2.15 - 5.(-2)3 + 4 = 2 - 5.(-8) + 4 = 2 + 40 + 4 = 46
Vậy ...
|\(x\) + 7| = 2
\(\left[\begin{array}{l}x+7=-2\\ x+7=2\end{array}\right.\)
\(\left[\begin{array}{l}x=-2-7\\ x=2-7\end{array}\right.\)
\(\left[\begin{array}{l}x=-9\\ x=-5\end{array}\right.\)
Vậy \(x\in\left(-9;-5\right)\)
Nếu \(x=-9\) thay vào biểu thức C = \(-\frac57x^2-\frac34x+1\) ta có:
C = - \(\frac57\left(-9\right)^2\) - \(\frac43\left(-9\right)\) + 1
C = - \(\frac57.81+12\) + 1
C = \(-\frac{405}{7}\) + \(\frac{84}{7}+\frac77\)
C = - \(\frac{321}{7}+\frac77\)
C = \(\) - \(\frac{314}{7}\)
Nếu \(x=-5\) thay vào biểu thức C = \(-\frac57x^2-\frac{4}{3x}+1\) ta có:
C = - \(\frac57.\left(-5\right)^2-\frac43.\left(-5\right)+1\)
C = - \(\frac57.25+\frac{20}{3}+1\)
C = \(-\frac{125}{7}+\frac{20}{3}+1\)
C = \(-\frac{375}{21}+\frac{140}{21}+\frac{21}{21}\)
C = - \(\frac{235}{21}+\frac{21}{21}\)
C = - \(\frac{214}{21}\)