20 x 29 =
Tìm x :
\(x^3\) + \(\dfrac{29}{28}\) = \(\dfrac{20}{29}\)
ổng hợp hai dao động điều hào cùng phương là?
A.
B.
C.
D.
Tính bằng cách thuận tiện nhất
45/100 + 9/20 x 29 + 45% x 30 + 0,9 x 20
\(\dfrac{45}{100}+\dfrac{9}{20}\times29+45\%\times30\) + 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
20 x X + 45 = 20 + 21 + 22 + ...... + 29
tìm X nha mọi ngươi em tick cho
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
11/11+20/20+(29/12-2/101)+111/1+x=
Nếu , và
, thì giá trị của
bằng bao nhiêu?
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(x-1)(x+1)(x+2)=24
b.\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
c.\(\dfrac{x-29}{30}+\dfrac{x-30}{29}=\dfrac{29}{x-30}+\dfrac{30}{x-29}\)
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 ...
\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)
\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)
tích 20 x 21 x .... x 29 có tận cùng bằng mấy chữ số 0?