a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)

mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)

nên \(x^2-10x-2000=0\)

\(\Leftrightarrow x^2+40x-50x-2000=0\)

\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)

\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)

Vậy: S={-40;50}

chịu

`(x^2-10x-29)/1971+(x^2-10x-27)/1973=(x^2-10x-1971)/1929+(x^2-10x-1973)/1927`

`<=>(x^2-10x-29)/1971-1+(x^2-10x-27)/1973-1=(x^2-10x-1971)/1929-1+(x^2-10x-1973)/1927-1`

`<=>(x^2-10x-200)/1971+(x^2-10x-200)/1973=(x^2-10x-200)/1971+(x^2-10x-200)/1927`

`<=>(x^2-10x-200)(1/1971+1/1973-1/1929-1/1927)=0`

`<=>x^2-10x-200=0` do `1/1971+1/1973-1/1929-1/1927<0`

`<=>x^2-20x+10x-200=0`

`<=>x(x-20)+10(x-20)=0`

`<=>(x-20)(x+10)=0`

`<=>` \(\left[ \begin{array}{l}x=20\\x=-10\end{array} \right.\) 

Vậy `S={20,-10}`

\(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5.\)

\(\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)\)\(=0\)

\(\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\left(50-x\right).\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

=> 50 - x = 0 \(\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)

=> x = 50

Ta có: \(A=\frac{108}{27}\cdot\frac{146}{29}-\frac{54}{27}\cdot\frac{202}{29}-\frac{16}{29}\)

\(=4\cdot\frac{146}{29}-2\cdot\frac{202}{29}-\frac{16}{29}\)

\(=\frac{584}{29}-\frac{404}{29}-\frac{16}{29}\)

\(=\frac{164}{29}\)

A

A

a) 27 x 35 + 73 x 6 - 27 x 29

= 27 x ( 35 - 29 ) + 73 x 6

= 27 x 6 + 73 x 6

= 6 x ( 27 + 73 )

= 6 x 100

= 600

b) 29 x ( 47 + 53 ) - 47 x ( 29 + 53 )

= 29 x 47 + 29 x 53 - 47 x 29 + 47 x 53

= 29 x ( 47 + 53 - 47 ) + 47 x 53

= 29 x 53 + 47 x 53

= 53 x ( 29 + 47 )

= 53 x 76

= 4028

c) 5 x 17 x 4 x 25 x 20

= ( 5 x 20 ) x ( 4 x 25 ) x 17

= 100 x 100 x 17

= 10000 x 17

= 170000

Mới lớp 3!

\(pt\Leftrightarrow\frac{29}{21}-\frac{x}{21}+\frac{27}{23}-\frac{x}{23}+\frac{25}{25}-\frac{x}{25}+\frac{23}{27}-\frac{x}{27}+\frac{21}{29}-\frac{x}{29}=-5\Leftrightarrow-x\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}\Leftrightarrow-x=\frac{-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}}{\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}}=-50\Leftrightarrow x=50\\ \Rightarrow S=\left\{50\right\}\)

     \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}-6=0\)

\(\Leftrightarrow\)\(\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1=0\) \(\Leftrightarrow\) \(\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1970}{29}+\frac{x-1999}{27}=0\)

\(\Leftrightarrow\)\(\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}\right)=0\)

Vì   \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}\ne0\)

\(\Rightarrow\)\(x-1999=0\)

\(\Leftrightarrow\)\(x=1999\)

Vậy...

Cảm ơn bạn nhiều nha. Lần sau mình có bài j khó nữa nhớ giúp mình vs nhá😊😊😊