tim x : x+2/5 =x-2/7
tim x:
7(2x-5)-5(7x-2)+2(5x-7)=(x-2)-(x+4)
x=?
7 \(\times\) ( 2\(x\) - 5) - 5 \(\times\) (7\(x\) - 2) + 2 \(\times\) (5\(x\) - 7) = (\(x\) - 2) - (\(x\) +4)
14\(x\) - 35 - 35\(x\) + 10 + 10\(x\) - 14 = \(x\) - 2 - \(x\) - 4
(14\(x\) - 35\(x\) + 10\(x\)) - (35 - 10+ 14) = -6
(- 21 \(x\) + 10\(x\)) - (25 + 14) = - 6
-11\(x\) - 39 = - 6
-11\(x\) = - 6 + 39
- 11\(x\) = 33
\(x\) = 33 : (-11)
\(x\) = - 3
bạn xem đề bài đúng chưa? Vế phải là phép nhân hay phép trừ?
2/5*x cong x*1/5=2/7.tim x
1 tim GTLN của M=x2+y2+7/x^2+y^2+5
2 tim đa thức f(x) biết f(x-1)=x^2-3x+5
1) \(M=\frac{x^2+y^2+7}{x^2+y^2+5}=1+\frac{2}{x^2+y^2+5}\)
Ta có: \(x^2+y^2\ge0,\forall x;y\)
=> \(x^2+y^2+5\ge5\) với mọi x; y
=> \(\frac{2}{x^2+y^2+5}\le\frac{2}{5}\)
=> \(M\le1+\frac{2}{5}=\frac{7}{5}\)
Dấu "=" xảy ra <=> x = y = 0
Vậy max M = 7/5 đạt tại x = y = 0
2) \(f\left(x-1\right)=x^2-3x+5=x^2-x-2x+2+3\)
\(=x\left(x-1\right)-2\left(x-1\right)+3=x\left(x-1\right)-\left(x-1\right)-\left(x-1\right)+3\)
\(=\left(x-1\right)\left(x-1\right)-\left(x-1\right)+3\)
=> \(f\left(x\right)=x.x-x+3=x^2-x+3\)
7(2x-5)-5(7x-2)+2(5x-7) = (x+2)-(x+4) tim x nhe ai giai giup minh vs
7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)
=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4
=> (14x - 35x + 10x) + (-35 + 10 - 14) = -2
=> -11x + (-39) = -2
=> -11x = -2 - (-39)
=> -11x = 37
=> x = \(\frac{-37}{11}\)
7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)
=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4
=> (14x - 35x + 10x) + (-35 + 10 - 14) = -6
=> -11x - 39 = -6
=> -11x = -6+39
=> -11x = 33
=> x = 33:(-11)
=> x = -3
tim x:(5x-7-9x):(-2-3x+7)=5-2(x-2)
tim x biet (x^2-1)(x^2-3)(x^2-5)(x^2-7)<=0
tim x
a, x-5)^4 =( x-5)^6
b,(x +2) ^5 =2^10
c,5^x :5^2 =125
d, (3x - 2^4) .7^3 =2 . 7^4
b) (x+2)5=210 c) 5x : 52 = 125 d) (3x-24) . 73= 2x74 ( câu a khá dài nên mình để sau nhé )
(x+2)5=45 5x : 52 = 53 (3x-24) . 73= 2x74 : 73
x+2=4 5x = 53 . 52 3x-24 = 2x7
x=4-2 5x = 55 3x -16 =14
x=2 x = 5 3x=14+16
3x=30
x=30:3
x=10
a) (x-5)4 = (x-5)6
(x-5)4 - (x-5)6 =0
(x-5)4 - (x-5)4 . (x-5)2 =0
(x-5)4 - [ 1- (x-5 )2 ] =0
TH1: ( x-5 )4 =0 TH2: 1-( x-5)2 = 0
x-5=0 (x-5)2 = 1-0
x=0+5 (x-5)2 =1
x=5 x-5 = +-1
* x-5=1
x=6
* x-5=-1
x=4
tim x: 1/5 x 7+1/7x9+...+1/X(x+2)=7/95
Đặt VT=A
\(2A=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}\)
\(2A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}\)
\(A=\left(\frac{1}{5}-\frac{1}{x+2}\right):2\)
Thay A vào VT ta đc:
\(\left(\frac{1}{5}-\frac{1}{x+2}\right):2=\frac{7}{95}\)
\(\frac{1}{5}-\frac{1}{x+2}=\frac{14}{95}\)
\(\frac{1}{x+2}=\frac{1}{19}\)
\(\Rightarrow x+2=19\)
\(\Rightarrow x=17\)
\(\text{Đặt VT=A }\)
\(2A=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}\)
\(2A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}\)
\(A=\left(\frac{1}{5}-\frac{1}{x+2}\right):2\)
\(\text{Thay A vào VT ta đc: }\)
\(\left(\frac{1}{5}-\frac{1}{x+2}\right):2=\frac{7}{95}\)
\(\frac{1}{5}-\frac{1}{x+2}=\frac{14}{95}\)
\(\frac{1}{x+2}=\frac{1}{19}\)
\(\Rightarrow x+2=19\)
\(\Rightarrow x=17\)
Tim x
2^x-1+5×2^x-2=7/32