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\(135-5\cdot\left(x+4\right)=35\)
\(5\cdot\left(x+4\right)=135-35\)
\(5\cdot\left(x+4\right)=100\)
\(x+4=100:5\)
\(x+4=20\)
\(x=20-4\)
\(x=16\)
b) \(25+3\cdot\left(x-8\right)=106\)
\(3\cdot\left(x-8\right)=106-25\)
\(3\cdot\left(x-8\right)=81\)
\(x-8=81:3\)
\(x-8=27\)
\(x=27+8\)
\(x=35\)
c) \(3^2\cdot\left(x+4\right)-5^2=5\cdot2^2\)
\(9\cdot\left(x+4\right)-25=5\cdot4\)
\(9\cdot\left(x+4\right)-25=20\)
\(9\cdot\left(x+4\right)=20+25\)
\(9\cdot\left(x+4\right)=45\)
\(x+4=45:9\)
\(x+4=5\)
\(x=5-4\)
\(x=1\)
Chúc bạn học tốt!!!!!!!!!!!!!!!!!!!!
a) 135 - 5 (x + 4) = 35
5(x+4)=135-35
5(x+4)=100
x+4=100:5
x+4=20
x=20-4
x=16
b)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=35
a,5(x+4)=135-35
5(x+4)=100
x+4=100:5
x+4=20
x=20-4
x=16
b,3(x-8)=106-25
3(x-8)=81
x-8=81:3
x-8=27
x=27+8
x=35
c) 135 - | 9 - x | = 3
| 9 - x | = 135 - 3
| 9 - x | = 132
=> 9 - x = 132
x = 9 - 132
x = 123
hoặc 9 - x = - 132
x = 9 - ( - 132 )
x = 9 + 132
x = 141
Vậy ..................
c) 135 - | 9 - x | = 35
\(2\left(x-1\right)+3\left(x-2\right)=x-4\)
\(2x-2+3x-6=x-4\)
\(5x-8=x-4\)
\(5x-8-x+4=0\)
\(4x-4=0\)
\(4x=4\)
\(x=1\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-x-2x+2=x+20\)
\(14-3x=x+20\)
\(14-3x-x-20=0\)
\(-6-4x=0\)
\(4x=-6\)
\(x=-\frac{3}{2}\)
\(\left|2x+3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}2x=2\\2x=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}}\)
a) 135 - 5(x + 4) = 35
<=> 135 - 5x - 20 = 35
<=> 115 - 5x = 35
<=> 5x = 115 - 35
<=> 5x = 80
<=> x = 16
b) 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 = .......
=a, \(\dfrac{x}{15}\) = \(\dfrac{2}{5}\)
= \(x.5=15.2\)
=> \(x=\dfrac{15.2}{5}\)\(=\dfrac{30}{5}\) \(=6\)
Vậy \(x=6\)
b, \(\dfrac{3}{x-7}\) \(=\dfrac{27}{135}\)
= \(\dfrac{3}{x-7}\) \(=\dfrac{3}{15}\)
= \(x-7=15\)
\(x=15+7\)
\(x=22\)
vậy x = 22
c, \(320.x-10=5.48:24\)
= \(320x-10=240:24\)
= \(320x-10=10\)
= \(320x=10+10\)
\(320x=20\)
\(x=20:320\)
\(x=0,0625\)
d, \(5x-1952=\) \(2500-1947\)
\(5x-1952=553\)
\(5x=553+1952\)
\(5x=2505\)
\(x=2505:5\)
\(x=501\)
e, \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)\left(x+5\right)=45\)
= \(\left(x+x+x+x+x\right)\)+\(\left(1+2+3+4+5\right)\) \(=45\)
= \(5x+15=45\)
\(5x=45-15\)
\(5x=30\)
\(x=30:5\)
\(x=6\)
f, \(x-\dfrac{2}{3}-\dfrac{2}{15}-\dfrac{2}{35}-\dfrac{2}{63}=\dfrac{1}{9}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}-\dfrac{2}{35}=\dfrac{1}{9}+\dfrac{2}{63}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}-\dfrac{2}{35}=\dfrac{1}{7}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}=\dfrac{1}{7}+\dfrac{2}{35}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}=\dfrac{1}{5}\)
= \(x-\dfrac{2}{3}=\dfrac{1}{5}+\dfrac{2}{15}\)
= \(x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\) \(\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=1\)
k, \(\dfrac{3+5+7+...+2015}{2+4+6+...+2014+x}=1\)
ta thấy phần tử là tập hợp các số lẻ ; phần mẫu là tập hợp các số chẵn
mà số chẵn hơn số lẻ 1 đơn vị
nên x thuộc tổng các số phần tử hơn mẫu là 1 đơn vị
=> từ \(2+4+6+...+2014\)có số số hạng là :
( 2014 - 2 ) : 2 + 1 = 1007
vậy x sẽ bằng :
( 1 + 1 ) . 1007 : 2 = 1007
vập số cần tìm là : 1007
a) \(4^{x^{ }}=64\)
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3.\)
b) Vì \(35⋮x\)
\(\Rightarrow x\inƯ\left(35\right)=\left\{1;5;7;35\right\}\)
Vậy \(x\in\left\{1;5;7;35\right\}\)
c) \(135-\left(x-5\right)=105:3\)
\(\Leftrightarrow\)\(135-\left(x-5\right)=35\)
\(\Leftrightarrow x-5=135-35\)
\(\Leftrightarrow x-5=100\)
\(\Leftrightarrow x=100+5\)
\(\Leftrightarrow\) \(x=105.\)
d) Vì \(x⋮25\)
\(\Rightarrow x\in B\left(25\right)=\left\{0;25;50;75;100;...\right\}\)
Mà \(x< 100\)
\(\Rightarrow x\in\left\{0;25;50;75\right\}\)
Dạng 2:
a) \(\left|x\right|=5\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
b) \(\left|x\right|< 2\) (vô lí, vì \(\left|x\right|\ge0\forall x\))
c) \(\left|x\right|=-1\) (vô lí, vì \(\left|x\right|\ge0\forall x\))
d) \(\left|x\right|=\left|-5\right|\Leftrightarrow\left|x\right|=5\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
e) \(\left|x+3\right|=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
f) \(\left|x-1\right|=4\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
g) \(\left|x-5\right|=10\Leftrightarrow\orbr{\begin{cases}x-5=10\\x-5=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=15\\x=-5\end{cases}}\)
h) \(\left|x+1\right|=-2\) (vô lí, vì \(\left|x+1\right|\ge0\forall x\))
i) \(\left|x+4\right|=5-\left(-1\right)\Leftrightarrow\left|x+4\right|=6\Leftrightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)
k) \(\left|x-1\right|=-10-3\Leftrightarrow\left|x-1\right|=-13\) (vô lí, vì \(\left|x-1\right|\ge0\forall x\))
l) \(\left|x+2\right|=12+\left(-3\right)+\left|-4\right|\Leftrightarrow\left|x+2\right|=13\Leftrightarrow\orbr{\begin{cases}x+2=13\\x+2=-13\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=11\\x=-15\end{cases}}\)
m) \(\left|x+2\right|-12=-1\Leftrightarrow\left|x+2\right|=11\Leftrightarrow\orbr{\begin{cases}x+2=11\\x+2=-11\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=-13\end{cases}}\)
n) \(135-\left|9-x\right|=35\Leftrightarrow\left|9-x\right|=100\Leftrightarrow\orbr{\begin{cases}9-x=100\\9-x=-100\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-91\\x=109\end{cases}}\)
\(\left|2x+3\right|=5\Leftrightarrow\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=2\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
a) 16 nha bn!!
b) 1 nha bn !!
k mik nha
a) 135 - 5 × ( x + 4 ) = 35
=>5x(X+4)=135-35
=>5x(x+4)=100
=>x+4=100:5
=>x+4=20
=>x=20-4
=>x=16
b) 3² × ( x + 4 ) - 5² = 5 × 2²
=>9x(x+4)-25=5x4
=>9x(x+4)-25=20
=>9x(x+4)=20+25
=>9x(x+4)=45
=>x+4=45:9
=>x+4=5
=>x=5-4
=>1