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1. a) (x-2)2 =1
=> x - 2 = \(\pm\sqrt{1}\)
=> x - 2 = 1 hoặc -1
=> x = 3 hoặc 1
b) 2x - 1= -8
=> 2x = -7
=>x = \(\dfrac{-7}{2}\)
c)thiếu đề
d) (x-1)x+2 = (x-1)x+4
(x-1)x+2 = (x-1)x+2+2
(x-1)x+2 = (x-1)x+2. (x-1)2
(x-1)x+2 - (x-1)x+2. (x-1)2 = 0
=> (x-1)x+2. [1 - (x-1)2] = 0
\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
=>-6a+5b=6a-5b
=>-12a=-10b
=>6a=5b
hay a/b=5/6
Xét 2 t.h là ra mà bn : a âm - b dương
a dương -b âm ( loại vì thế k thỏa mãn bài )
minhf cũng làm theo cach này nhưng cô bảo là chưa chắc đã dc điểm
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
x,y tỉ lệ thuận với \(\dfrac{3}{4}\) và \(\dfrac{4}{3}\)
\(\Rightarrow\dfrac{x}{\dfrac{3}{4}}=\dfrac{y}{\dfrac{4}{3}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :
\(\dfrac{x}{\dfrac{3}{4}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{x+y}{\dfrac{3}{4}+\dfrac{4}{3}}=-\dfrac{50}{\dfrac{25}{12}}=-24\)
\(\dfrac{x}{\dfrac{3}{4}}=-24\Rightarrow x=-18\)
\(\dfrac{y}{\dfrac{4}{3}}=-24\Rightarrow y=-32\)
Vì x tỉ lệ thuận với \(\dfrac{3}{4}\)\(\Rightarrow x=\dfrac{3}{4}.k\)
Vì y tỉ lệ thuận với \(\dfrac{4}{3}\Rightarrow y=\dfrac{4}{3}.k\)
\(\Rightarrow x+y=\dfrac{3}{4}.k+\dfrac{4}{3}.k\)
Mà x+y=50
\(\Rightarrow\dfrac{3}{4}.k +\dfrac{4}{3}.k=-50\)
\(\Rightarrow\left(\dfrac{3}{4}+\dfrac{4}{3}\right).k=-50\)
\(\Rightarrow\dfrac{25}{12}.k=-50\)
\(\Rightarrow k=-50:\dfrac{25}{12}\)
\(\Rightarrow k=-24\)
\(\Rightarrow x=\dfrac{3}{4}.\left(-24\right)=-18\)
Tick mk nha!!!
\(y=\dfrac{4}{3}.\left(-24\right)=-32\)
Vậy \(x=-18,y=-32\)
a,
\(\dfrac{1916\cdot1918-2}{1915+1916\cdot1917}\\ =\dfrac{1916\cdot\left(1917+1\right)-2}{1916\cdot1917+1915}\\ =\dfrac{1916\cdot1917+1916-2}{1916\cdot1917+1915}\\ =\dfrac{1916\cdot1917+1914}{1916\cdot1917+1915}\)
Vì \(1914< 1915\Rightarrow1916\cdot1917+1914< 1916\cdot1917+1915\Rightarrow\dfrac{1916\cdot1917+1914}{1916\cdot1917+1915}< 1\)
Vậy \(\dfrac{1916\cdot1917+1914}{1916\cdot1917+1915}< 1\)
b,
Áp dụng \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)
Ta có:
\(B=\dfrac{10^{22}+1}{10^{23}+1}< 1\\ \Rightarrow A=\dfrac{10^{22}+1}{10^{23}+1}< \dfrac{10^{22}+1+9}{10^{23}+1+9}=\dfrac{10^{22}+10}{10^{23}+10}=\dfrac{10\cdot\left(10^{21}+1\right)}{10\cdot\left(10^{22}+1\right)}=\dfrac{10^{21}+1}{10^{22}+1}=B\)
Vậy \(A< B\)
@Đời về cơ bản là buồn... cười!!! bạn giúp mik vs!