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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 \(x=-\dfrac{5}{2}\)
Chả hỉu gì cả.
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
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)\(5.3^x=7.3^5-2.3^5\)
\(\Rightarrow5.3^x=3^5\left(7-2\right)\)
\(\Rightarrow5.3^x=3^5.5\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Tìm x dễ thì tự làm nha:
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)-\left(\dfrac{x+2}{2002}+1\right)-\left(\dfrac{x+1}{2003}\right)=0\)\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\text{a) }\dfrac{\left(-6\right)^6}{216}=\dfrac{6^6}{216}=\dfrac{6^6}{6^3}=6^3=216\)
\(\text{b) }\dfrac{64}{\left(-4\right)^5}=-\dfrac{64}{4^5}=-\dfrac{4^3}{4^5}=-\dfrac{1}{4^2}=-\dfrac{1}{16}\)
\(\text{c) }\dfrac{900}{\left(-30\right)^3}=-\dfrac{900}{30^3}=-\dfrac{30^2}{30^3}=-\dfrac{1}{30}\)
\(\text{d) }\dfrac{225}{15^3}=\dfrac{15^2}{15^3}=\dfrac{1}{15}\)
a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)
\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)
= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)
= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)
\(\dfrac{51}{2.50}=\dfrac{51}{100}\)
Lời giải:
a)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow \left(\frac{a}{b}\right)^2=\left(\frac{b}{d}\right)^2=\frac{(a+c)^2}{(b+d)^2}(1)\)
Mặt khác, \(\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}(2)\) (áp dụng tính chất dãy tỉ số bằng nhau)
Từ \((1),(2)\Rightarrow \frac{(a+c)^2}{(b+d)^2}=\frac{a^2+c^2}{b^2+d^2}\)
b) Vì \(1-\frac{1}{2^2};1-\frac{1}{3^2};...;1-\frac{1}{50^2}<1\) nên:
\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)
\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)
\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)
\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)
\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)
1b. Ta thấy \(225-15^2=0\)
Mọi số nhân với 0 đều = 0
=> \(2017^0=1\)
2.
\(A=\dfrac{2.5^{22}-9.5^{21}}{25^{10}}:\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\dfrac{5^{21}\left(2.5-9\right)}{5^{20}}:\dfrac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}=5.1:\dfrac{5.7^{14}.2}{7^{15}.10}=5:\dfrac{1}{7}=35\)
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
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
\(12^n:2^{2n}=3^n.\left(2^2\right)^n:2^{2n}=3^n.2^{2n}:2^{2n}=3^n\)
\(3^8:3^4+2^2.2^3=3^4+2^5=81+32=113\)
\(\left(7^{1997}-7^{1995}\right)\left(7^{1994}\cdot7\right)=7^{1995}\left(7^2-1\right)\cdot7^{1995}=7^{1995\cdot2}\cdot48=7^{3990}\cdot48\)
\(4^{14}\cdot5^{28}=4^{14}\cdot\left(5^2\right)^{14}=\left(4\cdot25\right)^{14}=100^{14}\)
\(3\cdot4^2-2\cdot3^2=3\cdot2^4-2\cdot3^2=6\left(2^3-3\right)=6\cdot5=30\)
\(18^3:9^3=\left(18:9\right)^3=2^3=8\)
\(\left(2^8+8^3\right):\left(2^5\cdot2^3\right)=\left(2^8+2^9\right):2^8=\dfrac{2^8}{2^8}+\dfrac{2^9}{2^8}=1+2=3\)
\(16\cdot64\cdot8^2:\left(4^3.2^5.16\right)=2^4\cdot2^6\cdot2^6:\left(2^6\cdot2^5\cdot2^4\right)=2\)
\(5\cdot2^9\cdot6^{10}-7\cdot2^{29}\cdot27^6=5\cdot2^9\cdot2^{10}\cdot3^{10}-7\cdot2^{29}\cdot3^{18}=2^{19}\cdot3^{10}\left(5\cdot3^{10}-7\cdot2^{10}\cdot8\right)\)