2^x+2 + 2^x+2 + 2^x+1=224
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\(1,x^2+9-16y^2+6x\\ =\left(x^2+6x+9\right)-\left(4y\right)^2\\ =\left(x+3\right)^2-\left(4y\right)^2\\ =\left(x-4y+3\right)\left(x+4y+3\right)\\ 2,x^2-9+y^2+2xy\\ =\left(x^2+2xy+y^2\right)-9\\ =\left(x+y\right)^2-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ 3,x^2-4x+4-9y^2\\ =\left(x-2\right)^2-\left(3y\right)^2\\ =\left(x-3y-2\right)\left(x+3y-2\right)\\ 4,x^2-4xy+4y^2-81\\ =\left(x-2y\right)^2-9^2\\ =\left(x-2y-9\right)\left(x-2y+9\right)\\ 5,6x^2+6y^2-24+12xy\\ =\left(6x^2+12xy+6y^2\right)-24\\ =6\left[\left(x^2+2xy+y^2\right)-4\right]\\ =6\left[\left(x+y\right)^2-2^2\right]\\ =6\left(x+y-2\right)\left(x+y+2\right)\\ 6,9x^2-6x+1-25\\ =\left(3x-1\right)^2-5^2\\ =\left(3x-1-5\right)\left(3x-1+5\right)\\ =\left(3x-6\right)\left(3x+4\right)\)
7: \(x^2+4x+4-49y^2\)
\(=\left(x^2+4x+4\right)-49y^2\)
\(=\left(x+2\right)^2-49y^2\)
=(x+2+7y)(x+2-7y)
8: \(a^3+9a-ab^2-6a^2\)
\(=a\left(a^2-6a+9-b^2\right)\)
\(=a\left[\left(a-3\right)^2-b^2\right]\)
\(=a\left(a-3-b\right)\left(a-3+b\right)\)
9: \(8x^2-16x+8-32y^2\)
\(=8\left(x^2-2x+1-4y^2\right)\)
\(=8\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)
=8(x-1-2y)(x-1+2y)
10: \(4x^2-4x+1-81a^2\)
\(=\left(4x^2-4x+1\right)-81a^2\)
\(=\left(2x-1\right)^2-\left(9a\right)^2\)
=(2x-1-9a)(2x-1+9a)
11: \(x^2-6xy+9y^2-121\)
\(=\left(x^2-6xy+9y^2\right)-121\)
\(=\left(x-3y\right)^2-11^2=\left(x-3y-11\right)\left(x-3y+11\right)\)
12: \(12x^2-24x+12-3y^2\)
\(=3\left(4x^2-8x+4-y^2\right)\)
\(=3\left[\left(2x-2\right)^2-y^2\right]=3\left(2x-2-y\right)\left(2x-2+y\right)\)
\(P=\dfrac{\left(x-1\right)^2+3}{\left(x-1\right)^2+5}=\dfrac{\left(x-1\right)^2+5-2}{\left(x-1\right)^2+5}=1-\dfrac{2}{\left(x-1\right)^2+5}\)
\(\left(x-1\right)^2+5>=5\forall x\)
=>\(\dfrac{2}{\left(x-1\right)^2+5}< =\dfrac{2}{5}\forall x\)
=>\(-\dfrac{2}{\left(x-1\right)^2+5}>=-\dfrac{2}{5}\forall x\)
=>\(P=\dfrac{-2}{\left(x-1\right)^2+5}+1>=-\dfrac{2}{5}+1=\dfrac{3}{5}\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
\(Q=\dfrac{\left(2y+3\right)^2-3}{\left(2y+3\right)^2+4}=\dfrac{\left(2y+3\right)^2+4-7}{\left(2y+3\right)^2+4}=1-\dfrac{7}{\left(2y+3\right)^2+4}\)
\(\left(2y+3\right)^2+4>=4\forall y\)
=>\(\dfrac{7}{\left(2y+3\right)^2+4}< =\dfrac{7}{4}\forall y\)
=>\(-\dfrac{7}{\left(2y+3\right)^2+4}>=-\dfrac{7}{4}\forall y\)
=>\(Q=-\dfrac{7}{\left(2y+3\right)^2+4}+1>=-\dfrac{3}{4}\forall y\)
Dấu '=' xảy ra khi 2y+3=0
=>2y=-3
=>y=-3/2
\(F=\dfrac{\left(x-1\right)^2+5}{\left(x-1\right)^2+2}=\dfrac{\left(x-1\right)^2+2+3}{\left(x-1\right)^2+2}=1+\dfrac{3}{\left(x-1\right)^2+2}\)
\(\left(x-1\right)^2+2>=2\forall x\)
=>\(\dfrac{3}{\left(x-1\right)^2+2}< =\dfrac{3}{2}\forall x\)
=>\(F=\dfrac{3}{\left(x-1\right)^2+2}+1< =\dfrac{5}{2}\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\\ =1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)-\dfrac{1}{x+1}\\ =1-\dfrac{1}{x+1}\\ =\dfrac{x+1}{x+1}-\dfrac{1}{x+1}\\ =\dfrac{x}{x+1}\)
\(5^{2x-1}=125\\ =>5^{2x-1}=5^3\\ =>2x-1=3\\ =>2x=3+1\\ =>2x=4\\ =>x=\dfrac{4}{2}\\ =>x=2\)
Vậy: ...
52x - 1 = 125
52x - 1 = 53
2x - 1 = 3
2x = 3 +1
2x = 4
x = 4 :2
x = 2
a, 19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 26
= ( 19 + 21 ) + ( 18 + 22 ) + ( 17 + 23 ) + ( 16 + 24 ) + ( 14 + 26 )
= 40 + 40 + 40 + 40 + 40
= 40 x 5
= 200
Chúc bé học tốt
a.
19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 26
= (19 + 21) + (18 + 22) + (17 + 23) + ( 16 + 24) + (14 + 26)
= 40 + 40 + 40 + 40 + 40
= 40 x 5
= 200
câu b mk đã lm ở dưới r nhé
c) 64 x 4 + 18 x 4 + 9 x 8
= 64 x 4 + 18 x 4 + 9 x 2 x 4
= 4 x ( 64 + 18 + 9 x 2)
=4 x (82 + 18)
= 4 x 100
= 400
\(15,2-x+12=93\)
\(15,2-x=93-12\)
\(15,2-x=81\)
\(x=15,2-81\)
\(x=-65,8\)
Em có nhầm ko nhỉ, lớp 4 sao đã có số âm được
lớp 4 chưa học số thập phân và số âm, hình như em viết sai đề bài.
x.16 - x.4 - x = 2
x. (16 - 4 - 1) = 2
x. 11 = 2
x = 2. 11
x = 22
Vậy x = 22
x.16 - x.4 - x = 2
x.(16-4-1) = 2
x.11 = 2
x = 2:11
x = \(\dfrac{2}{11}\)
3 x 9 + 18 x 2 + 2 x 9 + 9
= 3 x 9 + 9 x 2 x 2 + 2 x 9 + 9
= 9 x (3 + 2 + 2 + 2 + 1)
= 9 x 10
= 90
\(=3\times9+9\times2\times2+2\times9+9\)
\(=3\times9+4\times9+2\times9+9\)
\(=9\times\left(3+4+2+1\right)\)
\(=9\times10\)
\(=90\)
C1:
\(A=\left\{0;1;2;3;4;5;6;7;8;9;10;11;12\right\}\)
\(C2:A=\left\{x\inℕ|x\le12\right\}\)
\(2^{x+2}+2^{x+2}+2^{x+1}=224\\ =>2^{x+1}\cdot2+2^{x+1}\cdot2+2^{x+1}\cdot1=224\\ =>2^{x+1}\cdot\left(2+2+1\right)=224\\ =>2^{x+1}\cdot5=224\\ =>2^{x+1}=\dfrac{224}{5}\\ =>x+1=log_2\dfrac{224}{5}\\ =>x=log_2\dfrac{224}{5}-1\)