Tìm x biết :
(12x - 4^3 ) . 8^10 = 4 . 8^11
tìm x biết a, 128 - 3 x ( x + 4 ) = 13 ; b , ( 12x - 4 ^ 3 ) x 8 ^3 = 4 x 8 ^ 4 ; c, [ ( 4x + 28 ) x 23 + 55 ] : 5 = 35
tìm x biết a, 128 - 3 x ( x + 4 ) = 13 ; b , ( 12x - 4 ^ 3 ) x 8 ^3 = 4 x 8 ^ 4 ; c, [ ( 4x + 28 ) x 23 + 55 ] : 5 = 35
a: =>3(x+4)=115
=>x+4=115/3
hay x=103/3
b: =>\(8^3\cdot\left(12x-64\right)=4\cdot8^4\)
\(\Leftrightarrow12x-64=32\)
=>12x=96
hay x=8
c: \(\Leftrightarrow\left(4x+28\right)\cdot23+55=175\)
=>(4x+28)x23=120
=>4x+28=120/23
=>4x=-524/23
hay x=-131/23
Tìm x biết:
1)25/12x+11/15=9/10
2)3/2x-2/5=1/3x-1/4
3)29/12[x]-5/6=3/8
4)[4x+3/4]-5/4=2
5)2x+2x+3=144
1) \(\frac{25}{12}.x+\frac{11}{15}=\frac{9}{10}\)
=> \(\frac{25}{12}.x=\frac{9}{10}-\frac{11}{15}\)
=> \(\frac{25}{12}.x=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{25}{12}\)
=> \(x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\).
3) \(\frac{29}{12}.\left[x\right]-\frac{5}{6}=\frac{3}{8}\)
=> \(\frac{29}{12}.\left[x\right]=\frac{3}{8}+\frac{5}{6}\)
=> \(\frac{29}{12}.x=\frac{29}{24}\)
=> \(x=\frac{29}{24}:\frac{29}{12}\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\).
4) \(\left[4x+\frac{3}{4}\right]-\frac{5}{4}=2\)
=> \(\left[4x+\frac{3}{4}\right]=2+\frac{5}{4}\)
=> \(4x+\frac{3}{4}=\frac{13}{4}\)
=> \(4x=\frac{13}{4}-\frac{3}{4}\)
=> \(4x=\frac{5}{2}\)
=> \(x=\frac{5}{2}:4\)
=> \(x=\frac{5}{8}\)
Vậy \(x=\frac{5}{8}\).
5) 2x + 2x+3 = 144
⇔ 2x + 2x . 23 = 144
⇔ 2x . (1 + 23) = 144
⇔ 2x . 9 = 144
⇔ 2x = 144 : 9
⇔ 2x = 16
⇔ 2x = 24
=> x = 4
Vậy x = 4.
Chúc bạn học tốt!
tìm x biết
a)3x3-12x=0
b)(x-3)2-(x-3)(3-x)2=0
c)4(x+1)+(2x+1)2-8(x-1)(x+1)=11
d)(x+3)(x2-3x+9)-x(x-4)(x+4)
a) 3x^3-12x=0
3x(x^2-4)=0
3x(x-2)(x+2)=0
suy ra 3x=0 suy ra x=0
x-2=0 x=2
x+2=0 x= -2
b) (x-3)^2-(x-3)(3-x)^2=0
(x-3)^2-(x-3)(x-3)^2=0
(x-3)^2(1-x+3)=0
(x-3)^2(4-x)=0
suy ra x-3=0 suy ra x=3
4-x=0 x=4
a) và b) đã nhé bạn
Tìm x, biết: (12x - 4)3.83 = 46.163
\(\left(12x-4\right)^3.8^3=4^{16}.16^3\)
\(\Leftrightarrow8\left(12x-4\right)^3=4^6.\left(4^2\right)^{^3}\)
\(\Leftrightarrow\left(96x-32\right)^3=4^6.4^6\)
\(\Leftrightarrow\left(96x-32\right)^3=16^6\)
\(\Leftrightarrow\left(96x-32\right)^3=\left(16^2\right)^{^3}\)
\(\Leftrightarrow\left(96x-32\right)^3=256^3\)
\(\Leftrightarrow96x-32=256\)
\(\Leftrightarrow x=\frac{256+32}{96}\)
Vậy \(x=3\)
(12x-4)3.83=43.163
<=> [(12x-4).8]3=(4.16)3
=> (12x-4).8=4.16
<=> 12x-4=8
<=> 12x=12
=> x=1
Tìm x , biết
a) \(\frac{8^{10}+4^{10}}{8^{11}+4^{11}}\)
b) x(x-2015)-2+2015=0
\(\frac{8^{10}+4^{10}}{8^{11}+4^{11}}=\frac{2^{30}+2^{20}}{2^{33}+2^{22}}=\frac{2^{20}.\left(2^{10}+1\right)}{2^{20}.\left(2^{13}+2^2\right)}=\frac{2^{10}+1}{2^{13}+4}=\frac{1025}{8196}\)
1/ Tính:M=(810+411) / (84+48)
2/ Tìm x biết: a, (x-3)2=25 b,(x-3)x-2=(x-3)x+2
bài 11: tìm x,y,z biết:
x+y+z=48
và \(12x-15y/7=20z-12x/9-19y-20z/11\)
b/
x/2=y/3=z/4 và 2x^2+3y^2-5z^2=-405
bài 12:
cho a/b=7/8;b/c=4/3
hỏi c tỉ lệ với a theo hệ số tỉ lậ mấy
Bài 12:
\(\dfrac{a}{b}=\dfrac{7}{8}\)
nên \(b=a:\dfrac{7}{8}=\dfrac{8}{7}a\)
Ta có: \(\dfrac{b}{c}=\dfrac{4}{3}\)
\(\Leftrightarrow b=\dfrac{4}{3}c\)
\(\Leftrightarrow a\cdot\dfrac{8}{7}=\dfrac{4}{3}c\)
\(\Leftrightarrow a=\dfrac{4}{3}:\dfrac{8}{7}\cdot c=\dfrac{4}{3}\cdot\dfrac{7}{8}\cdot c=\dfrac{7}{6}c\)
Vậy: c tỉ lệ với a theo hệ số tỉ lệ k=6/7
A= x^2 + 11x + 3
B= x^2 - 12x + 5
C= 3x^2 + 7 + 4
D= 7x^2 + 8x + 10
M= 16x^2 - 24x + 11
E= -3x^2 + 12x + 8
F= -25x^2 - 50x + 3