1978 x 2018 - 1956 phần 1978 x 2016 + 2000
giúp mình nhé
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cọc a ko bik làm :)))
b) 1980 x 1979 - 1978 x 1979
= ( 1980 - 1978 ) x 1979
= 2 x 1979
= 3958
Chúc bạn hok tốt!~~~
b) 1980 . 1979 - 1978 . 1979
= 1979 ( 1980 - 1978 )
=1979 . 2
= 3958
=))
1979 x 1979 + 1980 x 1958 / 1980 x 1979 - 1978 x 1979
= (1979x1979-1978x1979) + \(\frac{1980X1958}{1980X1979}\)
= (1979-1978) x 1979 + \(\frac{1958}{1979}\)
= 1 x 1979 x \(\frac{1958}{1979}\)
= 1979 x \(\frac{1958}{1979}\)
=\(\frac{1979X1958}{1979}\)
=\(1958\)
GHI RÕ TỪNG CHI TIẾT , CHÚC BẠN HỌC TỐT !!!!!
\(\frac{1978\times1999-1}{1978\times1998+1997}=\frac{1978\times1998+\left(1998-1\right)}{1978\times1998+1997}=\frac{1978\times1998+1997}{1978\times1998+1997}=1\)
a ) ( 1978 x 1979 + 1980 x 21 + 1958 ) / ( 1980 x 1979 - 1978 x 1979 ) = 1978 x 1979 + 1980 / 1979 . ( 1980 - 1978 )
= 0/1979
\(x^2=1978\)
\(=>x=\orbr{\begin{cases}-\sqrt{1978}\\\sqrt{1978}\end{cases}}\)
Các câu dễ tự làm :v
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1981}=-4\) (sau khi đã sửa đề)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1981}+1\right)=0\)\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1981}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1981}\right)=0\)
\(\Rightarrow2013-x=0\Rightarrow x=2013\)
\(1+5+9+13+17+.....+x=5050\)
Số các số hạng là:
\(\dfrac{x-1}{4}+1=\dfrac{1}{4}x+\dfrac{3}{4}\)
Như vậy có :
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\) số hạng
Theo đề bài ta có:
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\left(x+1\right)=5050\)
\(\Rightarrow\left(\dfrac{1}{4}x+\dfrac{3}{4}\right)\left(x+1\right)=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+\dfrac{1}{4}x+\dfrac{3}{4}x+\dfrac{3}{4}=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+x+\dfrac{3}{4}=10100\)
Kiệt sức.đến đây ko nghĩ nổi nx
a,
\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ 5^x=5^2\\ \Rightarrow x=2\)
Vậy \(x=2\)
b,
\(\left(x+2\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
Vậy \(x=7\) hoặc \(x=-11\)
d,
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\\ \dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}+4=0\\ \dfrac{45-x}{1968}+1+\dfrac{40-x}{1973}+1+\dfrac{35-x}{1978}+1+\dfrac{30-x}{1983}+1=0\\ \dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\\ \left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên
\(2013-x=0\\ x=2013\)
Vậy \(x=2013\)
e,
\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\\ 2015x=\dfrac{-2015}{2016}\\ x=\dfrac{-1}{2016}\)
Vậy \(x=\dfrac{-1}{2016}\)
\(\frac{1978x2018-1956}{1978x2016+2000}\)
=\(\frac{\left[1978x\left(2016+2\right)-1956\right]}{\left(1978x2016+2000\right)}\)
=\(\frac{1978x2016+1978x2-1956}{\left(1978x2016+2000\right)}\)
=\(\frac{\left(1978x2016+3956-1956\right)}{\left(1978x2016+2000\right)}\)
=\(\frac{\left(1978x2016+2000\right)}{\left(1978x2016+2000\right)}\)
= 1
hok tốt ~
\(x\)là dấu nhân nha