Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài1:
Giải 1 câu các câu sau tương tự
1.A=|x|+1
Với mọi x thì |x|>=0
=>|x|+1 >=1
Hay A>=1
Để A=1 thì |x|=0
=>x=0
Vậy...
Bài2:
1.A=−|x−2|+7
Với mọi x thì −|x−2|nhỏ hơn bằng 0
=>−|x−2|+7 nhỏ hơn bằng 7
Hay A nhỏ hơn bằng 7
Để A=7 thì |x−2|=0
=>x-2=0=>x=2
Các câu sau tương tự
1) \(A=\left|x\right|+1\ge1\forall x\)
\(\Rightarrow GTNN\) của A là 1 khi \(\left|x\right|=0\Leftrightarrow x=0\)
vậy GTNN của A là 1 khi \(x=0\)
2) \(B=\left|x+1\right|-\dfrac{7}{3}\ge-\dfrac{7}{3}\forall x\)
\(\Rightarrow GTNN\) của B là \(-\dfrac{7}{3}\) khi \(\left|x+1\right|=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
vậy GTNN của B là \(-\dfrac{7}{3}\) khi \(x=-1\)
3) \(C=\dfrac{2}{5}\left|2x+5\right|-2\ge-2\forall x\)
\(\Rightarrow GTNN\) của C là -2 khi \(\left|2x+5\right|=0\Leftrightarrow2x+5=0\Leftrightarrow2x=-5\Leftrightarrow x=-\dfrac{5}{2}\)
vậy GTNN của C là -2 khi \(x=-\dfrac{5}{2}\)
Cái này dễ lắm. Mình giải luôn nhé!
a) \(\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\Leftrightarrow x=\dfrac{2}{7}:\dfrac{1}{7}\Leftrightarrow x=2\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\Leftrightarrow x=-\dfrac{3}{5}:\left(-\dfrac{1}{5}\right)\Leftrightarrow x=3\\\dfrac{1}{3}x+\dfrac{4}{3}=0\Leftrightarrow x=-\dfrac{4}{3}:\dfrac{1}{3}\Leftrightarrow x=-4\end{matrix}\right.\)
Vậy x=2 hoặc x=3 hoặc x=-4
b)\(x\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)+1=0\)
\(x.0+1=0\)
\(1=0\) ( vô lí)
Vậy không có giá trị của x nào thỏa mãn
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
\(a,\left|x\right|+\left|x+2\right|=0\)
Với mọi x thì \(\left|x\right|\ge0;\left|x+2\right|\ge0\)
=>\(\left|x\right|+\left|x+2\right|\ge0\) với mọi x
Để \(\left|x\right|+\left|x+2\right|=0thì\)
\(x=0vàx=-2\)
=>\(x\in\varnothing\)
Vậy......
\(b,\left|x\left(x^2-\dfrac{5}{4}\right)\right|=0\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-\dfrac{5}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\pm\dfrac{\sqrt{5}}{4}\end{matrix}\right.\)
Vậy..
\(a,\left|x\right|+\left|x+2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x\right|=0\\\left|x+2\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\left(-2\right)\end{matrix}\right.\)
Mà \(0\ne\left(-2\right)\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
a: \(A=6\left(x+\dfrac{1}{3}\right)^2-7>=-7>-8\forall x\)
\(B=-8-\left(3.75-x\right)^2\le-8\)
Do đó: A>B
b: \(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}=\dfrac{15}{16}\)
\(B=\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}\)
Do đó: A>B
\(C=\left|x+1\right|+\left|x-2\right|+\left|x+3\right|\\ =\left|x+1\right|+\left(\left|2-x\right|+\left|x+3\right|\right)\\ \ge0+\left|2-x+x+3\right|\\ =5\)
Dấu "=" xảy ra khi \(\left(2-x\right)\left(x+3\right)\ge0\\ \)
\(\Rightarrow\left\{{}\begin{matrix}2-x\ge0\\x+3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2\\x\ge-3\end{matrix}\right.\Rightarrow-3\le x\le2\)
Vậy Min C = 5 khi \(-3\le x\le2\)
1. \(\left(-\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(-\dfrac{3}{2}\right)^x=\left(-\dfrac{3}{2}\right)^2\)
\(\Rightarrow x=2\)
2.\(3^{2x+2}=9^{10}\)
\(\Rightarrow3^{2x+2}=\left(3^2\right)^{10}\)
\(\Rightarrow3^{2x+2}=3^{20}\)
\(\Rightarrow2x+2=20\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
3)\(3^{3-2x}=27^{13}\)
\(\Rightarrow3^{3-2x}=\left(3^3\right)^{13}\)
\(\Rightarrow3^{3-2x}=3^{39}\)
\(\Rightarrow3-2x=39\)
\(\Rightarrow2x=-36\)
\(\Rightarrow x=-18\)
4)\(5.3^x=7.3^5-2.3^5\)
\(\Rightarrow5.3^x=3^5\left(7-2\right)\)
\(\Rightarrow5.3^x=3^5.5\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
\(12^n:2^{2n}=3^n.\left(2^2\right)^n:2^{2n}=3^n.2^{2n}:2^{2n}=3^n\)
\(3^8:3^4+2^2.2^3=3^4+2^5=81+32=113\)
\(\left(7^{1997}-7^{1995}\right)\left(7^{1994}\cdot7\right)=7^{1995}\left(7^2-1\right)\cdot7^{1995}=7^{1995\cdot2}\cdot48=7^{3990}\cdot48\)
\(4^{14}\cdot5^{28}=4^{14}\cdot\left(5^2\right)^{14}=\left(4\cdot25\right)^{14}=100^{14}\)
\(3\cdot4^2-2\cdot3^2=3\cdot2^4-2\cdot3^2=6\left(2^3-3\right)=6\cdot5=30\)
\(18^3:9^3=\left(18:9\right)^3=2^3=8\)
\(\left(2^8+8^3\right):\left(2^5\cdot2^3\right)=\left(2^8+2^9\right):2^8=\dfrac{2^8}{2^8}+\dfrac{2^9}{2^8}=1+2=3\)
\(16\cdot64\cdot8^2:\left(4^3.2^5.16\right)=2^4\cdot2^6\cdot2^6:\left(2^6\cdot2^5\cdot2^4\right)=2\)
\(5\cdot2^9\cdot6^{10}-7\cdot2^{29}\cdot27^6=5\cdot2^9\cdot2^{10}\cdot3^{10}-7\cdot2^{29}\cdot3^{18}=2^{19}\cdot3^{10}\left(5\cdot3^{10}-7\cdot2^{10}\cdot8\right)\)