Câu 24: (1đ) Tìm xeZ , biết :
a) 2x – 12 = -26 b) 15 – 5.( x – 3) = 4
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b ) 3 - ( x - 5 ) = - 7 + ( - 8 )
3 - ( x - 5 ) = - 15
x - 5 = 3 - ( - 15 )
x - 5 = 18
x = 18 + 5
x = 23
Vậy x = 23
Bài 1:
a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)
\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)
\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)
\(=\frac{-16}{7}\)
b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)
\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)
\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)
\(=\frac{10}{3}\)
Bài 2:
a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)
b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)
\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)
\(=1-1\frac{14}{15}=\frac{14}{15}\)
Bài 3:
a) x/5 = 2/5
=> x =2
b) -4/x = 20/14 = 10/7
=> -4/x = 10/7
=> x.10 = (-4).7
x.10 = - 28
x= -28 :10
x= -2,8
c) 4/7 = 12/x = 12/ 21
=> 12/x = 12/21
=> x = 21
d) 3/7 = x / 21 = 9/21
=> x/21 = 9/21
=> x= 9
a) ( x - 4 ) . ( x - 3 ) = 0
Thỏa mãn điều kiện x - 4 hoặc x - 3 bằng 0 .
\(\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}\)
b) ( x - 4 ) . ( y - 3 ) = 0
Thỏa mãn điều kiện x - 4 hoặc y - 3 bằng 0 .
\(\Rightarrow\hept{\begin{cases}x=4,x=3\\y=4,y=3\end{cases}}\)
c) x - 24 : 12 = 15
x - 2 = 15
x = 13
d) ( x - 12 ) : 13 = 5
x - 12 = 65
x = 77
e) 1 + 3 + 5 + ..... + x = 36
..... + x = 27
..... + x = 20
x = 11
f) 2 + 4 + .... + 2x = 110
2 + 4 + ....x = 55
Thỏa mãn điều kiện \(x\in N\left|2x+0\right|x⋮2\)
\(x\in\varnothing\)
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x-84=30x-240-6x+84\)
\(\Rightarrow\left(72-84\right)-\left(20x+36x\right)=\left(30x-6x\right)-240+84\)
\(\Rightarrow-12-56=24x-56x\)
\(\Rightarrow-12+156=24x+56x\)
\(\Rightarrow144=80x\)
\(\Rightarrow x=144:80\)
\(\Rightarrow x=\frac{9}{5}\)
b) \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x+25-8x+12-5x-6x-36-1=0\)
\(\Rightarrow-4x=0\)
\(\Rightarrow-4.0\)
\(\Rightarrow x=0\)
a, 2x - 12 = -26 => 2x = (-26) + 12 = -14 => x = (-14) : 2 = -7. b, 15 - 5.(x - 3) = 4 => 5.(x - 3) = 15 - = 11 (vô lý)(loại). Vậy n ∊ ∅. Hok tốt