tính va tim x:
a,36.72-36.67
b,{296:[78-(579-30.7):9]}.18
c,3x-5=16
d,42-2.(32-2x+1)=10
GIUP MINH
36.72-36.67
{296:[78-(579-30.7):9]}.18
3x-5=16
42-2.(32-2x+1)=10
GIÚP MÌNH,HELP MY
36.72-36.67=36.(72-67)=36.5=180 {296:[78-(579-30.7):9]}.18={296:[78-369:9]}.18=296:37.18=8.18=144 3x-5=16=3x-5=16+5.3x+5=21.x = 21:3.x = 7 vay x =7 phan kia minh ko bie,bn thong cam cho minh nha.minh can nhgieng cuu
\(36\cdot72-36\cdot67\)
\(=36\left(72-67\right)\)
\(=36\cdot5\)
\(=180\)
(296:(78-(579-30.7):9)).18
=(296:(78-(579-30.7):9)).18
=(296:(78-(579-210):9)).18
=(296:(78-369:9)).18
=(296:(78-41)).18
=(296:37).18
=8.18
=144
Tìm số nguyên x
a) 0100 - (31 - x) = 40
b) 280 - (x - 140) : 35 = 270
c) ( 1900 - 2x) : 35 - 32 = 16
d) 2^2x-1 : 4 = 8^3
e) (x + 2)^5 = 2^10
f) (3x - 4) . (x - 1)^3 = 0
g) (-2x + 1)^2 = 49
h) 1+2+3+...+x = 78
a: =>31-x=60
=>x=-29
b: =>(x-140):35=280-270=10
=>x-140=350
=>x=490
c: =>(1900-2x):35=48
=>1900-2x=1680
=>2x=220
=>x=110
d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)
=>2x-1=11
=>x=6
e: =>(x+2)^5=4^5
=>x+2=4
=>x=2
f: =>3x-4=0 hoặc x-1=0
=>x=4/3 hoặc x=1
g: =>(2x-1)^2=49
=>2x-1=7 hoặc 2x-1=-7
=>x=-3 hoặc x=4
h: =>x(x+1)/2=78
=>x(x+1)=156
=>x=12
Bài 8. Tìm x ϵ N biết:
a) x3=27
b) (2x-1)3=8
c) (x-2)2=16
d) (2x – 3)2=9
e) 2x+5=34:32
f) (3x-24).73=2.74
Bài 8. Tìm x ϵ N biết:
a) x3=27
b) (2x-1)3=8
c) (x-2)2=16
d) (2x – 3)2=9
e) 2x+5=34:32
f) (3x-24).73=2.74
tim x:a,(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1)=28
b,(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49
b)(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49
(=) x3- 6x2 +12 x -8 - ( x3 - 27 ) + 6( x2 + 2x +1)
(=) x3 - 6x2 +12x -8 - x3 +27 + 6x2 +12x +6
(=) 24x + 25 = 49
(=) 24x = 49 - 25 = 24
(=) x = 24/24 =1
Tìm x:
a)(3x-7)2=(2-2x)2
b)x2-8x+6=0
c)4x2-2x-1=0
d)x4-4x2-32=0
\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)
a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)
\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)
\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)
b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)
\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)
\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)
\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
c, \(4x^2-2x-1=0\)
\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)
\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)
\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
d,\(x^4-4x^2-32=0\)
đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)
\(< =>t^2-2.2t+4-6^2=0\)
\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)
\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)
bài 4: chứng minh biểu thức không phụ thuộc vào biến x:
a)M=(3x-5)(2x+11)-(2x+3)(3x+7)
b)N=(x+2)(2x2-3x+4)-(x2-1)(2x+1)
a)\(M=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+23x-55-6x^2-23x-21=-76\)
b) \(N=\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1=9\)
Tìm x:
a)x.(5-2x)-2x.(1-x)=15
b)(3x+2)^2+(1+3x).(1-3x)=2
a)x.(5-2x)-2x.(1-x)=15
x [ 5 - 2x -2.(1-x) ] = 15
x ( 5 - 2x -2 + 2x ) =15
x . 3 =15
x = 5
b)(3x+2)2+(1+3x).(1-3x)=2
9x2+12x+4+1-9x2=2
12x + 5 = 2
12x = -3
x = -1/4
a)\(\Leftrightarrow\)\(5x-2x^2-2x+2x^2=15\)
\(\Leftrightarrow\)\(3x=15\)
\(\Leftrightarrow\)\(x=5\)
b)\(\Leftrightarrow\)\(9x^2+12x+4+1-9x^2-2=0\)
\(\Leftrightarrow\)\(12x+3=0\)
\(\Leftrightarrow\)\(x=-0,25\)