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Ta có : \(\left|x\right|\ge0\forall x\in R\)
=> \(\left|x\right|+\frac{4}{7}\ge\frac{4}{7}\forall x\in R\)
=> GTNN của biểu thức là \(\frac{4}{7}\) khi x = 0
Ta có : |x - 2010| \(\ge0\forall x\in R\)
|x - 1963| \(\ge0\forall x\in R\)
Nên |x - 2010| + |x - 1963| \(\ge0\forall x\in R\)
Mà x ko thể đồng thời có 2 giá trị nên
GTNN của biểu thức là : 2010 - 1963 = 47 khi x = 2010 hoặc 1963
\(\left|x-2010\right|+\left|x-1963\right|=\left|x-2010\right|+\left|1963-x\right|\ge\left|x-2010+1963-x\right|=47\)
Dấu = xảy ra khi \(1963\le x\le2010\)
a/ có: \(\left|x+\dfrac{1}{5}\right|-x=x+\dfrac{1}{5}-x=\dfrac{1}{5}\)\(\forall x\)
=> \(A=\dfrac{1}{5}+\dfrac{4}{7}=\dfrac{27}{35}\)
=> A k có GTNN
b/ \(B=\left|x-2010\right|+\left|x-1963\right|=\left|x-2010\right|+\left|1963-x\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(\left|x-2010\right|+\left|1963-x\right|\ge\left|x-2010+1963-x\right|=\left|-47\right|=47\)
Đẳng thức xảy ra khi \(1963\le x\le2010\)
p/s: Đề a sai ak
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a: \(\Leftrightarrow\dfrac{39}{7}:\left\{x\cdot\dfrac{10}{13}+7.2\cdot\dfrac{257}{79}\right\}=\dfrac{15}{14}\)
\(\Leftrightarrow x\cdot\dfrac{10}{13}+\dfrac{9252}{395}=\dfrac{26}{5}\)
\(\Leftrightarrow x\simeq-23,69\)
b: TH1: x<1/2
Pt sẽ là 2-3x+1-2x=4
=>-5x+3=4
=>-5x=1
=>x=-1/5(nhận)
TH2: 1/2<=x<2/3
Pt sẽ là 2x-1+2-3x=4
=>1-x=4
=>x=-3(loại)
TH3: x>=2/3
Pt sẽ là 3x-2+2x-1=4
=>5x-3=4
=>5x=7
=>x=7/5(nhận)
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x+2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=19\)
Chúc bạn học tốt!!!
a, \(\dfrac{x+1}{5}+\dfrac{x+3}{4}=\dfrac{x+5}{3}+\dfrac{x+7}{2}\)
\(\Rightarrow\dfrac{x+1}{5}+2+\dfrac{x+3}{4}+2=\dfrac{x+5}{3}+2+\dfrac{x+7}{2}+2\)
\(\Rightarrow\dfrac{x+11}{5}+\dfrac{x+11}{4}-\dfrac{x+11}{3}-\dfrac{x+11}{2}=0\)
\(\Rightarrow\left(x+11\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Vậy x = -11
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
2) a) \(\left(x+\dfrac{4}{5}\right)^2=\dfrac{9}{25}\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{3}{5}\\x+\dfrac{4}{5}=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{5}\\x=\dfrac{-7}{5}\end{matrix}\right.\) vậy \(x=\dfrac{-1}{5};x=\dfrac{-7}{5}\)
b) \(\left|x-\dfrac{3}{7}\right|=-2\) vì giá trị đối không âm được nên phương trình này vô nghiệm
c) điều kiện : \(x\ge-7\) \(\sqrt{x+7}-2=4\Leftrightarrow\sqrt{x+7}=4+2=6\)
\(\Leftrightarrow x+7=6^2=36\Leftrightarrow x=36-7=29\) vậy \(x=29\)
d) \(x^2-\dfrac{7}{9}x=0\Leftrightarrow x\left(x-\dfrac{7}{9}\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-\dfrac{7}{9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{7}{9}\end{matrix}\right.\) vậy \(x=0;x=\dfrac{7}{9}\)
1) tìm GTNN
a) \(B=\left|x-2017\right|+\left|x-20\right|\)
B \(\ge\left|x-2017-x+20\right|=\left|-1997\right|=1997\)
Dấu " = " xảy ra khi và chỉ khi 20 \(\le x\le2017\)
Vậy MinB = 1997 khi 20 \(\le x\le2017\)
b) \(C=\left|x-3\right|+\left|x-5\right|\)
\(C\ge\left|x-3-x+5\right|=\left|2\right|=2\)
Dấu " = " xảy ra khi 3 \(\le x\le5\)
Vậ MinC = 2 khi và chỉ khi 3 \(\le x\le5\)
c) \(C=\left|x^2+4\right|+3\)
Ta thấy \(x^2+4\ge0\) với mọi x
nên \(\left|x^2+4\right|+3=x^2+4+3=x^2+7\)\(\ge\) 7
Dấu " =" xảy ra khi x = 0
MinC = 7 khi và chỉ khi x = 0
a) \(\left|x\right|=9,5\Leftrightarrow\left[{}\begin{matrix}x=9,5\\x=-9,5\end{matrix}\right.\)
b) \(\left|x+2\right|=\left|\dfrac{-3}{20}\right|=\dfrac{3}{20}\Leftrightarrow\left[{}\begin{matrix}x+2=\dfrac{3}{20}\\x+2=-\dfrac{3}{20}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{29}{15}\\x=-\dfrac{31}{15}\end{matrix}\right.\)
c) \(\left|x\right|=-2,4\Rightarrow x\in\varnothing\left(\left|x\right|\ge0\right)\)
d) \(\left|x+2,8\right|=1,5\Leftrightarrow\left[{}\begin{matrix}x+2,8=1,5\\x+2,8=-1,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1,3\\x=-4,3\end{matrix}\right.\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
a: \(=\left(\dfrac{11}{17}+\dfrac{6}{17}\right)+\left(-\dfrac{5}{13}-\dfrac{8}{13}\right)+\dfrac{11}{25}\)
=11/25+1-1=11/25
b: \(=\sqrt{36\cdot\dfrac{1}{4}}+11=9+11=20\)
c: \(=\left(0.25\right)^8\cdot4^8=\left(0.25\cdot4\right)^8=1\)
d: \(=2.8\left(-6.5-3.5\right)=-10\cdot2.8=-28\)
a/ \(\left|x\right|+\dfrac{4}{17}\)
Có: \(\left|x\right|\ge0\forall x\Rightarrow\left|x\right|+\dfrac{4}{17}\ge\dfrac{4}{17}\)
Dấu ''='' xảy ra \(\Leftrightarrow x=0\)
=> \(A_{MIN}=\dfrac{4}{17}\Leftrightarrow x=0\)
b/ \(\left|x+2,8\right|-6,9\)
Vì \(\left|x+2,8\right|\ge0\forall x\)
\(\Rightarrow\left|x+2,8\right|-6,9\ge-6,9\)
Dấu ''='' xảy ra \(\Leftrightarrow x=-2,8\)
Vậy \(B_{MIN}=-6,9\Leftrightarrow x=-2,8\)
bạn ghi kí tự lớp 7 đuo85c không khó hiểu quá