Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,x.\left(x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)
\(2,\left(x+12\right).\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)
\(3,\left(-x+5\right).\left(3-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}-x=-5\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
\(4,24:\left(3x-2\right)=-3\)
\(3x-2=-8\)
\(3x=-6\)
\(x=-2\)
\(5,-45:5\left(-3-2x\right)=3\)
\(5\left(-3-2x\right)=-15\)
\(-3-2x=-3\)
\(2x=0\)
\(x=0\)
\(6,x.\left(2+x\right)\left(7-x\right)=0\)
\(x=0\) hoặc \(\orbr{\begin{cases}2+x=0\\7-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=7\end{cases}}}\)
\(7,\left(x-1\right)\left(x+2\right)\left(-x+3\right)=0\)
TH1: x-1=0 TH2 : x+2=0 TH3: -x+3=0
x=1 x=-2 -x=-3 => x=3
a) 2x . 4 = 128
2x = 128 : 4
2x = 32
x = 32 : 2
x = 16
b)x . 17 = x
=> x = 0
Bài 1:
a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)
b) \(x^{10}=x\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)
\(\left(2x-15\right)^2=1\Rightarrow x=8\)
Bài 2:
\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)
\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)
C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)
\(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)
Vì 25<105 nên 19920<200315
\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)
Bài 1:
\(\frac{3}{5}+\frac{4}{15}=\frac{9}{15}+\frac{4}{15}=\frac{13}{15}\)
\(\frac{5}{6}:\frac{-7}{12}=\frac{5}{6}.\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
\(\frac{-21}{24}:\frac{-14}{8}=\frac{-21}{24}.\frac{-8}{14}=\frac{168}{336}=\frac{1}{2}\)
\(\frac{4}{5}:\frac{-8}{15}=\frac{4}{5}.\frac{-15}{8}=\frac{-60}{40}=\frac{-3}{2}\)
\(\frac{5}{12}-\frac{-7}{6}=\frac{5}{12}+\frac{7}{6}=\frac{5}{12}+\frac{14}{12}=\frac{19}{12}\)
\(\frac{-15}{16}.\frac{8}{25}=\frac{-120}{400}=\frac{-3}{10}\)
Bài 2 :
\(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)
\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)
\(=\left(\frac{34}{5}-\frac{19}{5}\right)-\frac{5}{3}\)
\(=3-\frac{5}{3}\)
\(=\frac{4}{3}\)
\(6\frac{5}{7}-\left(1\frac{2}{3}+2\frac{5}{7}\right)\)
\(=\frac{47}{7}-\left(\frac{5}{3}+\frac{19}{7}\right)\)
\(=\frac{47}{7}-\frac{5}{3}-\frac{19}{7}\)
\(=\left(\frac{47}{7}-\frac{19}{7}\right)-\frac{5}{3}\)
\(=4-\frac{5}{3}\)
\(=\frac{7}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+\frac{-3}{7}.\frac{15}{19}+\frac{5}{7}\)
\(=\left(\frac{4}{19}+\frac{15}{19}\right).\frac{-3}{7}+\frac{5}{7}\)
\(=1.\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{-3}{7}+\frac{5}{7}\)
\(=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
B1:
Ta có: \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}-2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.\left(-4\right)}{3.5}=-\frac{8}{15}\)
B2:
Ta có: \(1+3+5+...+x=1600\)
\(\Leftrightarrow\frac{\left(x+1\right)\cdot\left(\frac{x-1}{2}+1\right)}{2}=1600\)
\(\Leftrightarrow\left(x+1\right)\cdot\frac{x+1}{2}=3200\)
\(\Leftrightarrow\left(x+1\right)^2=6400\)
Xét theo dãy tăng tiến ta thấy được giá trị của x càng tăng
=> x dương => x + 1 dương
\(\Rightarrow x+1=80\)
\(\Rightarrow x=79\)
2) (x - 2)10 - (x - 2)8 = 0
(x - 2)2[(x - 2)2 - 1] = 0
x - 2 = 0 hoac (x - 2)2 - 1 = 0
x = 0 + 2 (x - 2)2 = 0 + 1
x = 2 (x - 2)2 = 1
x = 3
Vay: x = 2 hoac x = 3
Cau 2 bai 2 lam tuong tu
thanks