Tìm x biết:
a, (x + 24) × 3 = 75
b, (x - 34) ÷ 7 = 15
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a) x=-2
b)x=7
c)x=11 hay -11
d)không có giá trị
e)x=85
f)x=-55
g)x=0
****
a) 4x - 15 = -75 -x
4x+x = -75 + 15
5x = 60
x= 60: 5
=> x= 12
b) 3| x-7| = 21
|x-7|= 21:3
|x-7|=7
=> x-7 =7 hoặc x-7=-7
+) x-7=7
x=7+7=14
+) x-7=-7
x= -7+7=0
=> x=14 hoặc x=0
c) Áp dụng t/c phân số bằng nhau
=> x= \(\frac{-3.\left(-2\right)}{6}\)=\(\frac{6}{6}\)=1
Thay x=1 => y= \(\frac{\left(-2\right).\left(-18\right)}{1}\)=\(\frac{36}{1}\)=36
Thay y =36 => z=\(\frac{\left(-18\right).24}{36}\)=\(\frac{-432}{36}\)=-12
vậy (x,y,z)= (1;36;-12)
(câu d dài quá vs lại cx dễ nên bn tự lm nha mk chỉ giúp đến đây thôi)
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
a) 4x- 15 = -75 - x
=> 4x + x = -75 + 15
=> 5x = -60
=> x = -60 : 5
=> x = -12
b) 3(x - 7) = 21
=> x - 7 = 21 : 3
=> x - 7 = 7
=> x = 7 + 7
=> x = 14
c) \(-\frac{3}{6}=\frac{x}{-2}=\frac{-18}{y}=\frac{-z}{24}\)
Ta có: +) \(-\frac{3}{6}=\frac{x}{-2}\) => \(\frac{1}{-2}=\frac{x}{-2}\) => x = 1
+) \(\frac{-3}{6}=\frac{-18}{y}\) => \(\frac{-18}{36}=\frac{-18}{y}\) => y = 36
+) \(-\frac{3}{6}=\frac{-z}{24}\) = > \(\frac{-12}{24}=\frac{-z}{24}\) => -z = -12 => z = 12
d) \(-\frac{8}{3}+\frac{-1}{4}< x< \frac{-2}{7}+\frac{-5}{7}\)
=> \(\frac{-32}{12}-\frac{3}{12}< x< \frac{-2-5}{7}\)
=> \(\frac{-35}{12}< x< -1\)
=> x = -2 (x \(\in\)Z)
a) 4x - 15 = -75 - x
4x + x = -75 + 15
5x = -60
=> x = -60 : 5 = -12
a)
− 12 . x = − 15 . − 4 − 12 − 12 . x = 60 − 12 − 12 . x = 48 x = 48 : − 12 x = − 4
b)
− 9 . x + 3 = − 2 . − 7 + 16 − 9 . x + 3 = 14 + 16 − 9 . x + 3 = 30 − 9 . x = 30 − 3 − 9 . x = 27 x = 27 : − 9 x = − 3
c)
− 12 . x − 34 = 2 − 12 . x = 2 + 34 − 12 . x = 36 x = 36 : − 12 x = − 3
a)
− 12 . x = − 15 . − 4 − 12 − 12 . x = 60 − 12 − 12 . x = 48 x = 48 : − 12 x = − 4
b)
− 9 . x + 3 = − 2 . − 7 + 16 − 9 . x + 3 = 14 + 16 − 9 . x + 3 = 30 − 9 . x = 30 − 3 − 9 . x = 27 x = 27 : − 9 x = − 3
c)
− 12 . x − 34 = 2 − 12 . x = 2 + 34 − 12 . x = 36 x = 36 : − 12 x = − 3
a) (71+x)-(-24-x)+(-35-x)=0 b)x-34-[(15+x)-(23-x)]=0
71+x+24+x+-35-x =0 x-34-15-x-23+x =0
(x+x-x)+(71+24-35) =0 (x-x+x)+(-34-15-23) =0
x+60 =0 x+-72 =0
x =0-60 x =0-(-72)
x = -60 x = 72
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)