Tìm x: 2^7x phần 2^3x=64
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\(3x+36=-7x-64\)
\(10x=-100\)
\(x=-10\)
\(-5x-178=14x+145\)
\(-19x=323\)
\(x=-17\)
Bài dưới sai nha, giờ mới đúng nek
\(\frac{2^{7x}}{2^{3x}}=64\)
\(\frac{2^{3x+4x}}{2^{3x}}=64\)
\(\frac{2^{3x}.2^{4x}}{2^{3x}}=64\)
\(\frac{2^{3x}}{2^{3x}}.2^{4x}=64\)
⇒ 24x = 64
24x = 26
⇒ 4x = 6
x = \(\frac{6}{4}\)
NHỚ NHA BÀI HỒI NÃY SAI NHA - CHÚC BẠN HỌC TỐT t_t
(x-1)(2x^2-8)=0
\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)
3x^2-8x+5=0
áp dụng công thức bậc 2 ta có:
\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)
\(\Rightarrow x=\dfrac{5}{3};x=1\)
(7x-1).2x-7x+1=0
\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)
a) \(4x-3=11-3x\)
\(\Leftrightarrow4x+3x=11+3\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
Vậy .............
b) \(x^3-4x^2+3x=0\)
\(\Leftrightarrow x\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x\left(x^2-x-3x+3\right)=0\)
\(\Leftrightarrow x\left[x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
Vậy .................
P/s: câu c bn gõ lại dc ko
Đăng ít một thôi bạn :v
a) 3x - (3 - 2x) = 0
3x - 3 + 2x = 0
5x - 3 = 0
5x = 0 + 3
5x = 3
x = 3/5
b) (x + 2).3 - 4x.3 = 0
3.(x + 2) - 12.x = 0
3[x + 2 - (4x)] = 0
x + 2 - 4 = 0
-3x + 2 = 0
-3x = 0 - 2
-3x = -2
x = 2/3
c) (x - 2)(x - 4)(1 - 7x) = 0
x - 2 = 0 hoặc x - 4 = 0 hoặc 1 - 7x = 0
x = 0 + 2 x = 0 + 4 -7x = 0 - 1
x = 2 x = 4 -7x = -1
x = 1/7
d) 4x2 - 1/4 = 0
4x2 = 0 + 1/4
4x2 = 1/4
x2 = 1/4 : 4
x2 = 1/16
x2 = (1/4)2
x = 1/4 hoặc x = -1/4
e) -3x2 + 48 = 0
3x2 - 48 = 0
3x2 = 0 + 48
3x2 = 48
x2 = 48 : 3
x2 = 16
x2 = 42
x = 4 hoặc x = -4
g) 3(1/2 - 1/3x)3 - 1/9 = 0
3(1/2 - x/3)3 - 1/9 = 0
3(1/2 - x/3)3 = 0 + 1/9
3(1/2 - x/3)3 = 1/9
(1/2 - x/3)3 = 1/9 : 3
(1/2 - x/3)3 = 1/27
(1/2 - x/3)3 = (1/3)3
1/2 - x/3 = 1/3
-x/3 = 1/3 - 1/2
-x/3 = -1/6
-x = -1/6.3
-x = -3/6 = -1/2
x = -1/2
m) 4x3 + 5x4 = 0
x3(4 + 5x) = 0
x = 0 hoặc 4 + 5x = 0
x = 0 5x = 0 - 4
5x = -4
x = -4/5
h) -x3 + 1/64x = 0
-x3 + x/64 = 0
x/64 - x3 = 0
x(1/64 - x3) = 0
x = 0 hoặc 1/64 - x2 = 0
x = 0 -x2 = 0 - 1/64
-x2 = -1/64
x2 = 1/64 = -+1/8
k) (x2 + 1)2 + 3x(x2 + 1) + 2 = 0
x4 + 2x2 + 1 + 3x3 + 3x + 2 = 0
x4 + 2x2 + 3 + 3x3 + 3x = 0
(x3 + 2x2 + 3)(x + 1) = 0
Mà x3 + 2x2 + 3 # 0 nên
x + 1 = 0
x = -1
1, x^2 +5x+4x+20
= x(x+4)+5(x+4)
= (x+5).(x+4)
2,x^2 -x -20
= x^2 -5x+4x-20
= x(x-5)+4(x-5)
=(x+4).(x-5)
3,3x^2 +3x -2x -2
= 3x(x+1)-2(x+1)
= (3x-2)(x+1)
4, 2x^2-4x-x-2
=2x(x-2)-x-2
=(2x-1)(x+2)
5,6x^2-2x+9x-3
=2x(3x-1)+3(3x-1)
=(2x+3)(3x-1)
6,3x^2 +2x+9x+6
=3x(x+3)+2(x+3)
=(3x+2)(x+3)
7,= 2(x^2-10x+3)
8,=x^4 +4^3
mk làm luôn ko chép đề nha bn
\(x^2+9x+20\)
\(\Leftrightarrow x^2+4x+5x+20\)
\(\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\)
\(x^2-x-20\)
\(\Leftrightarrow x^2-5x+4x-20\)
\(\Leftrightarrow\left(x^2-5x\right)+\left(4x-20\right)\)
\(\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)\)
\(3x^2+x-2\)
\(\Leftrightarrow3x^2+3x-2x-2\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(2x+2\right)\)
\(\Leftrightarrow3x\left(x+1\right)-2x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-2x\right)\)
\(\)
\(\frac{2^{7x}}{2^{3x}}=64\)
\(\Rightarrow\frac{\left(2^7\right)^x}{\left(2^3\right)^x}=2^6\)
\(\Rightarrow\left(\frac{128}{8}\right)^x=2^6\)
\(\Rightarrow16^x=2^6\)
\(\Rightarrow2^{4x}=2^6\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\frac{6}{4}=\frac{3}{2}\)
Ta co 27x / 23x = 64
===> 27x-3x = 64
===> 24x = 64 = 26
=====> 4x = 6
===> x = 1,5