\(3^x\) + \(3^x\) + 1 = 36
chứng minh B = 36 / 1 x 3 x 5 + 36 / 3 x 5 x7 + 36 / 5 x 7 x 9 + .... + 36 / 25 x 27 x29 < 3
\(B=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=9.\left(\frac{1}{3}-\frac{1}{675}\right)\)
\(=9.\frac{224}{675}\)
\(=\frac{224}{75}\)
THIẾU
\(=\frac{224}{75}< \frac{225}{75}=3\)
Vậy \(B< 3\)
mấy bọn nguuuuuuuuuuuuuuuuuu
1, Tìm x, biết \(x^2\) – 36 = 0
A. x = 6. B. x = -6.
C. x = 6; x = -6. D. x = 36 hoặc x = - 36.
2, Tìm x, biết \(x^3\) – 3\(x^2\) + 3x - 1 = 0
A. x = 1. B. x = -1. C. x = 0. D. x = 2.
\(\dfrac{3}{4}\) x 36 + 75% x 23 + \(\dfrac{75}{100}\) + 0,75 x 40
= ..... x 36 + ....... x 23 + ..... + 0,75 x 40
= ..... x ( 36 + 23 + 1 + ..... )
= ..... x 100
= .....
1) Tìm x thuộc Z :
a) (x^2-3).(x^2-36) = 0
b) (x^2-3).(x^2-36) < 0
a) \(\left(x^2-3\right)\left(x^2-36\right)=0\)
TH1: \(x^2-3=0\Rightarrow x^2=3\)
Ta thấy không có số nguyên nào mà bình phương nên bằng 3 nên không có giá trị x thỏa mãn.
TH2: \(x^2-36=0\Rightarrow x^2=36=6.6=\left(-6\right).\left(-6\right)\)
Vậy x = 6 hoặc x = -6.
b) \(\left(x^2-3\right)\left(x^2-36\right)< 0\)
Do \(x^2-3>x^2-36\) nên chỉ có thể xảy ra trường hợp \(\hept{\begin{cases}x^2-3>0\\x^2-36< 0\end{cases}}\)
\(\Rightarrow3\le x^2\le36\Rightarrow2\le x\le6\) hoặc \(-6\le x\le-2\)
Bài1 thực hiện phép tính
a 5/2x^2+6x - 4-3x^2/x^2-9 -3
b , 3x^2+5x+14/x^3+1 + x-1/x^2-x+1 - 4/x+1
c, x-6/x^2+1 × 3x^2-3x+3/x^2-36 + x-6/x^3+1 × 3x/x^3-36
d,x^2+1/3x ÷ x^2+1/x-1 ÷x^3-1/x^2+x ÷ x^2+2x+1/x^2+x+1
tìm x biết:
a. 2.3^x - 405=3^(x-1)
b. (3/4)^x = 2^8/3^4
c. (5x+1)^2=36/49
d. (x+1)^(x+10)= (x+1)^(x+4)
Tim x la phan tu cua N biet
a ) ( x - 5 ) x 4 = 36
b) ( 48 - x ) x 2 = 14
c) ( 2 x X - 3 ) x 5 = 4
d) 1 + 3 + 5 +..........+ X = 36
e) 1 + 2 + 3 + ...........+ X = 55 + 6 +.....+ 2X = 110
c, (2\(\times\) \(x\) - 3) \(\times\) 5 = 4
2 \(\times\) \(x\) - 3 = \(\dfrac{4}{5}\)
2 \(\times\) \(x\) = 0,8 + 3
2\(x\) = 3,8
\(x\) = \(3,8\) : 2
\(x\) = 1,9
d, 1 + 3 + 5 +....+\(x\) = 36
(\(x\) + 1)\(\times\) \(x\): 2 = 36
(\(x+1\)) \(\times\)\(x\) = 72
(\(x+1\))\(\times\)\(x\) = 8 x 9
\(x\) = 8
a) (X - 5) x 4 = 36
⇒ X - 5 = 36 : 4
⇒ X - 5 = 9
⇒ X = 9 + 5
⇒ X = 14
Vậy X = 14
b) (48 - X) x 2 = 14
⇒ 48 - X = 14 : 2
⇒ 48 - X = 7
⇒ X = 48 - 7
⇒ X = 41
Vậy X = 41
Tìm x:
3^x + 3^x+1 = 36
3^x(1+ 3 ) = 36
=>4.3^x = 36
=> 3^x = 36 : 4
=> 3^x = 9
=> 3^x = 3^2
=> x= 2
tìm x
b) (x + 3)2 - (x - 4)( x + 8) = 1;
c) 3(x + 2)2 + (2x - 1)2 - 7(x + 3)(x - 3) = 36;
d)(x - 3)(x2 + 3x + 9) + x(x + 2)(2 - x) = 1;
e) (x + 1)3 - (x - 1)3 - 6(x - 1)2 = -19.
hihi
b)(x+3)2-(x-4)(x+8)=1
\(\Rightarrow\)x2+6x+9-(x2+8x-4x-32)=1
⇒x2+6x+9-x2-8x+4x+32=1
⇒2x+41=1
\(\Rightarrow\)2x+41-1=0
\(\Rightarrow\)2x+40=0
⇒2x=-40
\(\Rightarrow\)x=\(\dfrac{-40}{2}\)
⇒x=-20
Tìm x:
a) 3.(x-1)^2-3x.(x-5)=1
b) 3.(x+2)^2+(2x-1)^2-7.(x-3)(x+3)=36
3(x−1)^2−3x(x−5)=1
⇒3(x^2−2x+1)−3x^2+15x=1
⇒3x^2−6x+3−3x^2+15x=1
=9x+3=1
⇒9x=(−3)+1
⇒x=−2/9
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Rightarrow3\left(x^2-2x+1\right)-3x^2+15x=1\)
\(\Rightarrow3x^2-6x+3-3x^2+15x=1\)
\(=9x+3=1\)
\(\Rightarrow9x=\left(-3\right)+1\)
\(\Rightarrow x=\frac{-2}{9}\)