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mình viết đáp án luôn nhé
a,x=845/528
b,-7/5
c,-47/140
d,43/1710
chúc bạn làm bài thành công nhe!!@
các bạn nhớ chọn k đúng cho mình nhé
\(\left(x+\frac{5}{3}\right)\left(x-\frac{5}{4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{5}{3}\right)=0\\\left(x-\frac{5}{4}\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{5}{4}\end{cases}}}\)
\(\frac{1}{3}+\frac{1}{2}\div x=-4\)
\(\frac{1}{2}\div x=-4-\frac{1}{3}\)
\(\frac{1}{2}\div x=-\frac{13}{3}\)
\(x=-\frac{3}{26}\)
a, (10/3:x).(-5/4)=-10/3
10/3:x=-10/3:(-5/4)
10/3:x=8/3
x=10/3:8/3
x=5/4
b,(-6/5+x):(-18/5)=-1/4
-6/5+x=-1/4.(-18/5)
-6/5+x=9/10
x=9/10-(-6/5)=9/10+6/5
x=21/10
c,-22/15.x+1/3=-2/5
-22/15.x=-2/5-1/3=-11/15
x=-11/15:(-22/15)
x=11/21
d,(0,25-30%x).1/3=-31/6+1/4=-59/12
1/4-3/10x=-59/12:1/3=-59/4
3/10x=1/4-(-59/4)=1/4+59/4=15
x=15:3/10
x=50
e,(0,5x-3/7):1/2=8/7
1/2x=8/7.1/2=4/7
x=4/7:1/2
x=8/7
Tìm x , biết :
|x-2| = -3
Vô lí vì |x-2| phải lớn hơn hoặc bằng 0
=> không có giá trị x nào thỏa mãn đề bài trên .
|x-7| - 5 + 3 = 1
<=> |x-7| - 5 = -2
<=> |x-7| = 3
<=> x - 7 = 3 hoặc x - 7 = -3
<=> x = 10 hoặc x = 4
Vậy x€ { 10 ; 4 }
|x+5| - 5 = 4 - (-3)
<=> |x+5| - 5 = 7
<=> |x+5| = 7 + 5
<=> |x +5| = 12
<=> x + 5 = 12 hoặc x + 5 =-12
<=> x = 7 hoặc x = -17
Vậy x € { 7 ; -17 }
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
\(-7\left(5-x\right)-2\left(x-10\right)=15\)
\(\Leftrightarrow-35+7x-2x+20=15\)
\(\Leftrightarrow5x-5=15\)
\(\Leftrightarrow5x=20\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
\(4\left(x-1\right)-3\left(x-2\right)=-|-5|\)
\(\Leftrightarrow4x-4-3x+6=-5\)
\(\Leftrightarrow x+2=-5\)
\(\Leftrightarrow x=-7\)
Vậy \(x=-7\)
~ học tốt ~
\(\dfrac{4}{5}\div\left(x+\dfrac{1}{15}\right)+\dfrac{5}{3}=\dfrac{1}{6}\)
⇔ \(\left(x+\dfrac{1}{15}\right)+\dfrac{5}{3}=\dfrac{1}{6}\cdot\dfrac{4}{5}\)
⇔ \(\left(x+\dfrac{1}{15}\right)+\dfrac{5}{3}=\dfrac{2}{15}\)
⇔ \(x+\dfrac{1}{15}=\dfrac{2}{15}-\dfrac{5}{3}\)
⇔ \(x+\dfrac{1}{15}=-\dfrac{23}{15}\)
⇔ \(x=-\dfrac{23}{15}-\dfrac{1}{15}\)
⇔ \(x=-\dfrac{8}{5}\)