1. 30.(x+2) = 100+6.( x-5) +24x
2.(x-3).(7-x) > 0
3.(x-3).(x-7) < 0
giải giúp em nha!!!!!!
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1) (x - 35) - 120 = 0
x - 35 = 120
x = 120 + 35
x = 155
2) 310 - (118 - x) = 217
118 - x = 310 - 217
118 - x = 93
x = 118 - 93
x = 25
3) 156 - (x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
4) 814 - (x - 305) = 712
x - 305 = 814 - 712
x - 305 = 102
x = 102 + 305 = 407
5) 100 - 7 - (x - 5) = 58
x - 5 = 93 - 58
x - 5 = 35
x = 35 + 5 = 40
6) 12(x - 1) : 3 = 43 + 23
4(x - 1) = 72
x - 1 = 18
x = 18 + 1 = 19
7) 24 + 5x = 75 : 73
24 + 5x = 49
5x = 25
x = 25 : 5 = 5
8) 5(x - 1) : 3 = 43 + 23
\(\dfrac{5}{3}\left(x-1\right)=72\)
x - 1 = \(\dfrac{216}{5}\)
x = 221/5
9) 5(x - 4)2 - 7 = 13
5(x - 4)2 = 20
(x - 4)2 = 4
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
10) (x + 1) + (x + 2) + ... + (x + 30) = 795
=> (x + x + x + ... + x) + (1 + 2 + 3 +...+ 30) = 795 (1)
Đặt A = 1 + 2 + 3 +...+ 30
Số số hạng trong A là: (30 - 1) : 1 + 1 = 30 (số)
Tổng A bằng : (30 + 1).30 : 2 =465
Thay A = 465 vào (1) , ta được:
30x + 465 = 795
=> 30x =330
=> x =11
1: =>x-35=120
=>x=120+35=155
2: =>118-x=310-217=93
=>x=118-93=25
3: =>x+61=156-82=74
=>x=74-61=13
4: =>x-305=814-712=102
=>x=102+305=407
5: =>93-(x-5)=58
=>x-5=35
=>x=40
6: =>4(x-1)=64+8=72
=>x-1=18
=>x=19
7: =>5x+24=49
=>5x=25
=>x=5
8: =>5(x-1):3=4^3+2^3=64+8=72
=>5(x-1)=216
=>x-1=216/5
=>x=221/5
1) 5.(3-x)+2.(x-7)=-14
15-5x+2x-14=-14
1-3x=-14
3x=15
X=5
2) 30.(x+2)-6.(x-5)-24x=100
30x+60-6x+30-24x=100
0X+90=100
0X=10 vô lí
=> ko có giá trị x thỏa mãn điều kiện
3) (3x-9)^2=36
3x-9=6
3x-9=-6
TH1:3x-9=6 TH2:3x-9=-6
3x=15 3x=3
X=5 x=1
Vậy….
4) (1-2x)^3=-27
(1-2x)3=(-3)3
1-2x=-3
2x=4
X=2
Vậy…
5) (x-3).(x-2)<0
=>x-3 và x-2 cùng dấu
TH1:x-3>0 TH2:x-3<0
x-2<0 x-2>0
=>X>3 =>x<3
X<2 x>2
=>x>3 =>x<2
Vậy 3<x<2
câu 6 chịuuuu
câu 5 hơi khó ko bt có đúng hay ko đâu :)))
3) \(\left(3x-9\right)^2=36\Leftrightarrow\orbr{\begin{cases}3x-9=6\\3x-9=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=15\\3x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)
4) \(\left(1-2x\right)^3=-27\)
<=> 1-2x=-3
<=> 3x=4
<=> \(x=\frac{4}{3}\)
5) (x-3)(x-2)<0
=> x-3 và x-2 trái dấu nhau
thấy x-3<x-2 => \(\hept{\begin{cases}x-3< 0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x>2\end{cases}\Leftrightarrow}2< x< 3}\)
6) làm tương tự
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
a) Ta có: 4x-20=0
⇔4x=20⇔4x=20
hay x=5
Vậy: S={5}
b) Ta có: 2x+x+12=02x+x+12=0
⇔3x+12=0⇔3x+12=0
⇔3x=−12⇔3x=−12
hay x=-4
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}\)
⇒ \(x=\dfrac{21}{52}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)
\(\dfrac{x}{3}=\dfrac{11}{21}\)
⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)
⇒ \(x=\dfrac{11}{7}\)
c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)
\(\dfrac{-17}{7}< x< -1\)
⇒ \(-17< x< -7\)
⇒ \(x\in\left\{-16;-15,....;-6\right\}\)
d) \(\dfrac{1}{6}+\dfrac{2}{5}\)
\(=\dfrac{5}{30}+\dfrac{12}{30}\)
\(=\dfrac{17}{30}\)
e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)
\(=\dfrac{12}{20}+\dfrac{-35}{20}\)
\(=\dfrac{-23}{20}\)
f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)
\(=\dfrac{4}{13}+\dfrac{-2}{5}\)
\(=\dfrac{20}{65}+\dfrac{-26}{65}\)
\(=\dfrac{-6}{65}\)
g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)
\(=\dfrac{-6}{58}+\dfrac{16}{58}\)
\(=\dfrac{10}{58}\)
h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)
\(=\dfrac{1}{5}+\dfrac{-4}{5}\)
\(=\dfrac{-3}{5}\)
j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)
\(=\dfrac{-2}{9}+\dfrac{5}{9}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)