Sử dụng MTBT 

chọn C

     \(\dfrac{45}{100}+\dfrac{9}{20}\times29+45\%\times30\) + 0,9 \(\times\) 20

=    \(\dfrac{9}{20}\) + \(\dfrac{9}{20}\times29\) + \(\dfrac{9}{20}\) \(\times\) 30 + 18

= \(\dfrac{9}{20}\) \(\times\) ( 1 + 29 + 30) + 18

= \(\dfrac{9}{20}\) \(\times\) 60 + 18

= 27 + 18

= 45

Ta có \(VP=\frac{\left(20+29\right)\times10}{2}=245\)

=> 20xX+45=245

20xX=200

X=10

20.x + 45 = 20 + 21 + 22 + ... + 29

=> 20.x + 45 = (29 + 20).[(29 - 20) : 1 + 1] : 2

=> 20.x + 45 = 49 . 10 : 2

=> 20.x + 45 = 245

=> 20.x = 245 - 45

=> 20.x = 200

=> x = 200 : 20

=> x =10

20 . x + 45 = 20 + 21 + 22 + .... + 29  ( 1 )

Đặt A = 20 + 21 + 22 + .... + 29

Số số hạng của A là :

( 29 - 20 ) : 1 + 1 = 10 ( số hạng )

Tổng của A là :

( 29 + 20 ) . 10 : 2 = 245  ( 2 )

Thay ( 2 ) vào ( 1 ) ta có :

20 . x + 45 = 245

=> 20 . x = 200

x = 200: 2 = 10

Vậy x = 10

ta có y+4=(x-2)²=x²-4x+4 (1)

x+4=(y+2)²=y²-4y+4 (2)

Cộng (1)và (2), vế theo vế ta có :

x+y+8=x²-4x+4+y²-4y+4

\(\Rightarrow\) x²+y²=5x+5y

a.

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(a=x^2+x-1\) , ta có pt:

\(\left(a+1\right)\left(a-1\right)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)

*Với a = 5 ta được:

\(x^2+x-1=5\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

*Với a = -5 ta được:

\(x^2+x-1=-5\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)

Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)

c)(ĐKXĐ: x khác 30;29)

\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)

\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)

\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)

\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)

\(\Leftrightarrow1741-59x=0\)

\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)

Vậy S={0;\(\dfrac{1741}{59}\);59}

b)(ĐKXĐ:x khác 2;3;4;5;6)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-6\right)\left(x-2\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=32\)

\(\Leftrightarrow x^2-8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow x=-2\) or x=10(đều thỏa)

Vậy ...

\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)

\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)

 3 chữ số 0