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\(40-2x=18\\ \Leftrightarrow2x=40-18\\ \Leftrightarrow2x=22\\ \Leftrightarrow x=22:2\\ \Leftrightarrow x=11\)
\(3x+1=27\\ \Leftrightarrow3x=27-1\\ \Leftrightarrow3x=26\\ \Leftrightarrow x=\dfrac{26}{3}\)
\(3x-5=25\\ \Leftrightarrow3x=25+5\\ \Leftrightarrow3x=30\\ \Leftrightarrow x=30:3\\ \Leftrightarrow x=10\)
1) \(A=\frac{7}{10\times11}+\frac{7}{11\times12}+\frac{7}{12\times13}+...+\frac{7}{69\times70}\)
\(A=7\times\left(\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+...+\frac{1}{69\times70}\right)\)
\(A=7\times\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\times\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\times\frac{3}{35}\)
\(A=\frac{3}{5}\)
2) \(B=\frac{1}{25\times27}+\frac{1}{27\times29}+\frac{1}{29\times31}+...+\frac{1}{73\times75}\)
\(B=\frac{1}{2}\times\left(\frac{2}{25\times27}+\frac{2}{27\times29}+\frac{2}{29\times31}+...+\frac{2}{73\times75}\right)\).
\(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\times\frac{2}{75}\)
\(B=\frac{1}{75}\)
3) \(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)
\(C=\frac{4}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2008\times2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=2\times\frac{502}{1005}\)
\(C=\frac{1004}{1005}\)
_Chúc bạn học tốt_
1 . Để số tự nhiên 2x98y chia hết cho 2,5 thì y = 0
Theo như dấu hiệu chia hết đã học , số có tổng chữ số chia hết cho 3 thì chia hết cho 3
Tổng các chữ số trong số đó là :
2 + 9 + 8 + 0 = 19
Vậy để số 2x980 chia hết cho 3 thì x = 5
Tổng của các chữ số nếu x = 5 là :
2 + 5 + 9 +8 + 0 = 24
Mà 24 chia hết cho 3 nên x = 5
Vậy số x = 5 ; y = 0
a) x − 3 49 = − 2 7 ⇒ x − 3 49 = − 14 49 ⇒ x = − 11
b) x 4 = x + 1 8 ⇒ 2 x 4 = x + 1 8 ⇒ x = 1
Lời giải:
$x^6-x^4+x^2+m=x^4(x^2-1)+(x^2-1)+m+1$
$=(x^2-1)(x^4+1)+m+1$. Như vậy, đa thức này chia cho $x^2-1$ dư $m+1$
Vì $x^6-x^4+x^2+m$ chia hết cho $x^2-1$ nên $m+1=0$
$\Leftrightarrow m=-1$
Đáp án B.
a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a. \(\dfrac{3}{4}\) + \(\dfrac{-4}{5}\) - \(\dfrac{1}{2}\) = \(\dfrac{-1}{20}\) - \(\dfrac{1}{2}\) = \(\dfrac{-11}{20}\)
b. (4 - \(\dfrac{5}{12}\) ): 2 + \(\dfrac{5}{24}\)
= \(\dfrac{43}{12}\) : 2 + \(\dfrac{5}{24}\)
= \(\dfrac{43}{24}\) + \(\dfrac{5}{24}\)
=\(\dfrac{48}{24}\) = 2
2.2
\(\dfrac{4}{7}\) .x - \(\dfrac{2}{3}\) = \(\dfrac{1}{5}\)
\(\dfrac{4}{7}\) x = \(\dfrac{1}{5}\) + \(\dfrac{2}{3}\)
\(\dfrac{4}{7}\) . x = \(\dfrac{13}{15}\)
x = \(\dfrac{91}{60}\)
2.1
\(a)\dfrac{3}{4}+\dfrac{-4}{5}-\dfrac{1}{2}\\ =\dfrac{3\times5}{4\times5}+\dfrac{-4\times4}{5\times4}-\dfrac{1\times10}{2\times10}\\ =\dfrac{15}{20}+\dfrac{-16}{20}-\dfrac{10}{20}\\ =\dfrac{15-16-10}{20}\\ =\dfrac{-11}{20}\)
\(b)\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{4\times12}{1\times12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48}{12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48-5}{12}\right):2+\dfrac{5}{24}\\ =\dfrac{43}{12}:2+\dfrac{5}{24}\\ =\dfrac{43}{12}\times\dfrac{1}{2}+\dfrac{5}{24}\\ =\dfrac{43}{24}+\dfrac{5}{24}\\ =\dfrac{43+5}{24}\\ =\dfrac{48}{24}\\ =2\)
x4-1=27
=>x4-1=33
=>x3=33(vì 4-1=3)
=> x=3
3x-1=27
=>3x-1=33
=>x-1=3
=>x=2
a) Ta có: x4-1 = 27
Mà 33 = 27
x3 = 33
Vậy x = 3
b) Ta có: 3x-1 = 27
Mà 33 = 27
=> 3x-1 = 33
=> x - 1 = 3
=> x = 3+1=4
Vậy x = 4