Tim x
2^x-1+5×2^x-2=7/32
bài 1:tính
(5+1/7)x2/3
bài 2:tìm x
x:4/7=21/32
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AI TK MÌNH MÌNH TK LẠI 3 LẦN.
Bài 1 : \(\left(5+\frac{1}{7}\right)\times\frac{2}{3}\)
\(=\frac{36}{7}\times\frac{2}{3}\)
\(=\frac{24}{7}\)
Bài 2 : \(x:\frac{4}{7}=\frac{21}{32}\)
\(\Rightarrow x=\frac{21}{32}\times\frac{4}{7}\)
\(\Rightarrow x=\frac{3}{8}\)
B1 : (5+1/7) x 2/3
= 36/7 x 2/3
= 24/7
B2 : x : 4/7 = 21/32
x = 21/32 : 4/7
x = 147/128
Bài 1. Tìm x, biết
b) -2x + 36 = 6
a) -5.x + 32 = (-2)3
d) êx - 4 ê< 7
f) 40 < 31 + êx ê< 47
g) | x + 3| ≤ 2
e) (x + 9) . (x2 – 25) = 0
h) (x – 5)2 = 9
Bài 1:
a) Ta có: \(-5x+32=\left(-2\right)^3\)
\(\Leftrightarrow-5x+32=-8\)
\(\Leftrightarrow-5x=-40\)
hay x=8
Vậy: x=8
b) Ta có: \(-2x+36=6\)
\(\Leftrightarrow-2x=6-36=-30\)
hay x=15
Vậy: x=15
e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
b,-2x+36=6
tương đương -2x=-30
tương đương x=15
a, -5x+32=(-2)^3
tương đương -5x+32=8
tương đương -5x=-24
tương đương x=24/5
1)tim x biet rang:
a)3^x-1=1/243
b)2^x+2^x+3=144
c)81^-2x.27x=9^5
2)tim tiep so ghang thu 5 cua day so sau:-1/a^2;2/a^3;-6/a^4;24/a^5;...
3)tim so tu nhien x biet :
a)4^x+4^x+3=4160
b)2^x-1+5.2^x-2=7/32
1) 27,56 x 35 + 27,56 x67- 27,56 x2
2) 4 x 35 x 25 x 5 x2
3) ( 32 x 18 - 16 x 36 ) x( 1+3+5+7+......+1001)
4) 3/5 + 2/5 x 1/2 - 2/3 x 3/4
5) 27 x 13 + 35 x 27 - 12 x 9
6) 33/5 x 15/11 + 22/27 x 81/44 - 3/5 : 9/5
1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)
= (964 + 1846,52) - 55,12
=2810,52 - 55,12
= 2755,4
2).( 4x 35 ) x ( 25 x 5 ) x 2
= ( 140 x 125 ) x2
= 17500 x 2
=35000
4). 3/10
5). 1188
6). 61/6
1,2755,4
2,35000
3,0
4,\(\frac{3}{10}\)
5,1188
6,\(\frac{61}{6}\)
Cau 1:
Tim x, biet: 1-4+7-10+.............-x=-75
Cau 2:
Cho x1, x2, x3, x4, x5 thuộc Z
Biết x1+ x2 + x3 + x4 + x5=0
và x1 + x2=x3+ x4= x4 + x5 =2
Tinh x3, x4 , x5
Cau 3: Tim x biet
(x+7+1) chia het cho (x+7)
tim x
2.x2-1/7.x=0
tim he so cua x2 trong khai trien x(x+1)3+x2(x-2)5
a ( -o, 75+ 1/2) ;4/3
b 7/5 , 7/4 -32/5
tim x
10/3,x + 67/4= -13,25
1) a) \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}\)
\(=\left(-0,75+\frac{1}{2}\right)\cdot\frac{3}{4}\)
\(=\left(-0,75+0,5\right)\cdot0,75\)
\(=-0,25\cdot0,75\)
\(=-0,1875\)
Vậy: \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}=-0,1875\)
b) \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{128}{20}\)
\(=-\frac{79}{20}=-3\frac{19}{20}\)
Vậy: \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}=-3\frac{19}{20}\)
2) \(\frac{10}{3}\cdot x+\frac{67}{4}=-13,25\)
<=> \(\frac{10}{3}\cdot x+\frac{67}{4}=-\frac{53}{4}\)
<=> \(\frac{10}{3}\cdot x=-\left(\frac{53}{4}+\frac{67}{4}\right)\)
<=> \(\frac{10}{3}\cdot x=-30\)
<=> \(x=-30:\frac{10}{3}\)
<=> \(x=-30\cdot\frac{3}{10}\)
<=> \(x=-9\)
Vậy: \(x=-9\)
Bai 1 : Tim x , biet :
a) (32 - x ) + (-7 +I x I ) - (I x I + 25 ) = 25
32-x-7+I x I - 25 = 25 ( bước này cô mk làm hộ )
b) 1 + { -2 - [ -3 +( -4 + I x I)]} = 1-[2-(-3-4)]
mai mk học rồi , giúp nha, I x I là giá trị tuyệt đối của x !!!