Tìm x:
a) x x 4 = 36 b) 70 : x = 7
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\(x.y=-5\)
\(\Leftrightarrow x.y=-5=-1.5=1.\left(-5\right)=5.\left(-1\right)=-5.1\)
th1\(\orbr{\begin{cases}x=-1\\y=5\end{cases}}\)
th2\(\orbr{\begin{cases}x=1\\y=-5\end{cases}}\)
th3\(\orbr{\begin{cases}x=5\\y=-1\end{cases}}\)
th4\(\orbr{\begin{cases}x=-5\\y=1\end{cases}}\)
1.
75+7+125+93
=(75+125)+(7+93)
=200+100
=300
4x36x25x50
=(4x25)x(36x50)
=100x1800
=180000
2.
x+23=123
x=123-23
x=100
vậy x=100
70-(x.8)=27
x.8=70-27
x.8=43
x=43:8
x=43/8
vậy x=43/8
a) số số hạng là :
(100-41):1+1=60 (số hạng)
b) số số hạng là :
(100-10):2+1=46 (số hạng)
Tìm x
700:7-3*x=36*25:9-70
100-3*x=100-70
100-3*x=30
3*x=70
x=70/3
Câu đầu ko rõ đề lắm nên làm câu 2 ha:
\(\frac{2222}{333}+\frac{111111}{151515}-\frac{2}{5}\)
\(=\frac{2222}{333}+\left(\frac{11}{15}-\frac{6}{15}\right)\)
\(=\frac{2222}{333}+\frac{1}{3}\)
\(=\frac{2222}{333}+\frac{111}{333}=\frac{2333}{333}\)
a) x ⋮ 12;x ⋮ 25; x ⋮ 30 ⇒ x ∈ BC(12,25,30)
12 = 22 . 3 ; 25 = 52 ; 30 = 5.2.3
BCNN(12,25,30) = 22 . 52 . 3 = 300
BC(12,25,30) = B(300) = {0,300,600,...}
Mà theo đề bài 0 ≤ x ≤ 500 ⇒ x = 0;300
b) 70 ⋮ x ; 84 ⋮ x ; 120 ⋮ x ⇒ x ∈ ƯC(70,84,120)
70 = 2.3.7 ; 84 = 22.3.7 ; 120 = 23 .5.3
ƯCLN(70,84,120) = 2
ƯC(70,84,120) = Ư(2) = {1,2}
Mà x ≥ 8 ⇒ x = ∅
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
Lớp 4?
a) \(\dfrac{x+1}{4}=\dfrac{36}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=144\)
\(\Rightarrow\left[{}\begin{matrix}x+1=12\\x+1=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-13\end{matrix}\right.\)
Vậy: \(x\in\left\{11;-13\right\}\)
b) \(\dfrac{x}{7}=\dfrac{55x-4}{28}\)
\(\Rightarrow4x=55x-4\)
\(\Rightarrow-51x=-4\)
\(\Rightarrow x=\dfrac{4}{51}\)
Vậy: \(x=\dfrac{4}{51}\)
a) \(\dfrac{x + 1}{4} = \dfrac{36}{x + 1} \)
\(\Rightarrow\) \(( x + 1 )( x + 1 ) = 36 . 4 \)
\(\Rightarrow ( x + 1 )^2 = 144 \)
\(\Rightarrow ( x + 1 )^2 = 12^2 = ( -12 )^2 \)
\(\Rightarrow\) \(x + 1 ∈ \) { \(12 ; -12 \) }
\(\Rightarrow \) \(x \) \(∈ \) { \(11 ; -13 \) }
Vậy \(x ∈ \) { \(11 ; -13 \) }
a) x x 4 = 36 b) 70 : x = 7
x = 36 : 4 x = 70 : 7
x = 9 x = 10