\((x-1)^3=125\)
\(2^{x+2}-2^x=96\)
\(\left(2x+1\right)^3=343\)
\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)\(16^x< 128^4\)
K
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Những câu hỏi liên quan
ND
Nguyễn Đăng Nhân
CTVHS
20 tháng 9 2023
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
AH
Akai Haruma
Giáo viên
20 tháng 9 2023
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
14 tháng 7 2016
\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-\left(2x-5\right)\right]=8.5\)
\(720:\left[41-\left(2x-5\right)\right]=40\)
\(41-\left(2x-5\right)=720:40\)
\(41- \left(2x-5\right)=18\)
\(2x-5=41-18\)
\(2x-5=23\)
\(2x=23+5\)
\(2x=28\)
\(x=28:2\)
\(\Rightarrow x=14\)
HS
2
26 tháng 10 2019
Bài 1: 25 + 3(x - 8) = 106
3(x - 8) = 106 - 25
3(x - 8) = 81
(x - 8) = 81 : 3
(x - 8) = 27
x = 27 + 8
x = 25
Bài 2: 720 : [41 - (2x - 5)] = 23 . 5
720 : [41 - (2x - 5)] = 8 . 5
720 : [41 - (2x - 5)] = 40
[41 - (2x - 5)] = 720 : 40
[41 - (2x - 5)] = 18
(2x - 5) = 41 - 18
(2x - 5) = 23
2x = 23 + 5
2x = 28
x = 28 : 2
x = 14
9 tháng 5 2019
a, \(3^2\left(x+14\right)-5^2=5.4\)
\(\Rightarrow9\left(x+14\right)-25=20\)
\(\Rightarrow9x+126=20+25\)
\(\Rightarrow9x+126=45\)
\(\Rightarrow9x=45-126\)
\(\Rightarrow9x=-81\Rightarrow x=-9\)
Vậy x = 9
S
9 tháng 5 2019
a,\(3^2\left(x+14\right)-5^2=5.4\)
\(\Rightarrow9\left(x-14\right)-25=20\)
\(\Rightarrow9\left(x-14\right)=20+25=45\)
\(\Rightarrow x-14=45:9=5\)
\(\Rightarrow x=5+14\)
\(\Rightarrow x=19\)
Vậy x = 19
b,\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-2x+5\right]=40\)
\(\Rightarrow41+5-2x=720:40\)
\(\Rightarrow46-2x=18\)
\(\Rightarrow2x=46-18=28\)
\(\Rightarrow x=28:2\)
\(\Rightarrow x=14\)
Vậy x = 14
~Study well~
22 tháng 4 2020
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
NN
22 tháng 4 2020
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
MH
19 tháng 2 2022
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
19 tháng 2 2022
1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
A
11 tháng 9 2020
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
A
11 tháng 9 2020
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
5 tháng 3 2021
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
[X-1]^3=125
[x-1]^3=5^3
x-1=5
x=5+1
x=6
Vậy x=6
a) \(\left(x-1\right)^3=125\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
b) \(2^{x+2}-2^x=96\)
\(\Rightarrow2^x.2^2-2^x=96\)
\(\Rightarrow2^x.3=96\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
c) \(\left(2x+1\right)^3=343=7^3\)
\(\Rightarrow2x+1=7\)
\(\Rightarrow x=3\)