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Ta có :
\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2-\frac{1}{64}=0\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1}{64}\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1^2}{8^2}\)
\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)
\(\Leftrightarrow\)\(\frac{3}{4}x-\frac{1}{2}=\frac{1}{8}\)
\(\Leftrightarrow\)\(\frac{3}{4}x=\frac{1}{8}+\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{3}{4}x=\frac{5}{8}\)
\(\Leftrightarrow\)\(x=\frac{5}{8}:\frac{3}{4}\)
\(\Leftrightarrow\)\(x=\frac{5}{8}.\frac{4}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{2}.\frac{1}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\)
Chúc bạn học tốt ~
\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2-\frac{1}{64}=0\)
\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2=0+\frac{1}{64}=\frac{1}{64}\)
\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)
=>\(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{8}\)
\(\frac{3}{4}.x=\frac{1}{8}+\frac{1}{2}\)
\(\frac{3}{4}.x=\frac{5}{8}\)
\(x=\frac{5}{8}:\frac{3}{4}\)
\(x=\frac{5}{6}\)
2/3.x + 1/4 = 7/12
2/3.x = 7/12 - 1/4
2/3.x = 1/3
x = 1/3 : 2/3
x = 1/2
Bài làm
\(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\frac{2}{3}x=\frac{4}{12}\)
\(\frac{2}{3}x=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{2}{3}\)
\(x=\frac{1}{3}.\frac{3}{2}\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
câu a+b dùng quy tắc chuyển vế
c, 3.(1/2-x)-5.(x-1/10)=-7/4
=>(3.1/2-3x)-(5x-5.1/10)=-7/4
=>3/2-3x-5x+1/2=-7/4
=>(3/2+1/2)-(3x+5x)=-7/4
=> 2-8x=-7/4
=>8x=15/4
=>x=15/4:8
=>x=15/32
a) 2.(1/4 - 3x) = 1/5 - 4x
=> 1/2 - 6x = 1/5 -4x
=> -6x + 4x = 1/5 - 1/2
=> -2x = -3/10 = 3/20
b) 4.(1/3 - x) + 1/2 = 5/6 +x
=> 4/3 - 4x + 1/2 = 5/6 +x
=> -4x - x = 5/6 - 4/3 - 1/2
=> -5x = -1
=> x= 1/5
c) 3. (1/2 - x) -5. ( x - 1/10) = -7/4
=> 3/2 - 3x - 5x + 1/2 = -7/4
=> -3x - 5x = -7/4 - 3/2 - 1/2
=> -8x = -15/4
=> x = 15/32
\(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-5\\x=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\in\varnothing\\x=5\end{cases}}}\)
Vậy x=5
TK MK ĐÊ RỒI MK LÀM TIẾP!
\(x-\frac{1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x=\frac{8}{3}+\frac{1}{9}\)
\(\Leftrightarrow x=\frac{24+1}{9}=\frac{25}{9}\)
\(a,A=\frac{x-4}{x+1}=\frac{(x+1)-1-4}{x+1}=1-\frac{5}{x+1}\)
Để \(x\in Z\)thì \(x+1\inƯ(5)\)
mà \(Ư(5)=(5;1;-1;-5)\)
Ta có bảng sau
x + 1 | 5 | 1 | -1 | -5 |
x | 4 | 0 | -2 | -6 |
Vậy \(x=(4;0;-2;-6)\)
\(b,B=\frac{3x-5}{x-2}=\frac{3x-6+1}{x-2}=\frac{3x-6}{x-2}+\frac{1}{x-2}=\frac{3(x-2)}{x-2}+\frac{1}{x-2}=3+\frac{1}{x-2}\)
Để \(x\in Z\)thì \(x-2\inƯ(1)\)
mà \(Ư(1)=(1;-1)\)
Với \(x-2=1\Rightarrow x=3\)
Với \(x-1=-1\Rightarrow x=0\)
Vậy \(x=(3;0)\)
Chúc bạn học tốt nhé
\(A=\frac{x-4}{x+1}=\frac{x+1-5}{x+1}=\frac{-5}{x+1}\)
\(\Rightarrow x+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta lập bảng :
x + 1 | 1 | -1 | 5 | -5 |
x | 0 | -2 | 4 | -6 |
Vì \(x\inℤ\)thì x ta tìm đc tm
\(B=\frac{3x+5}{x-2}=\frac{3\left(x-2\right)+11}{x-2}=\frac{11}{x-2}\)
\(\Rightarrow x-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
Ta lập bảng :
x - 2 | 1 | -1 | 11 | -11 |
x | 3 | 1 | 13 | -9 |
Vì x\(\inℤ\)nên x ta tìm đc tm
(x+1)+(x+2)+(x+3)+...+(x+1024)=2056
<=> (x+x+x+...+x)+(1+2+3+...+1024)=2056
Số số hạng x là : (1024-1):1+1=1024 số
Đặt A = 1+2+3+..+1024
Tổng A là : (1024+1)x1024:2=524800
=> 1024x + 524800= 2056
<=> 1024x= 2056-524800
<=> 1024x=- 522744
x= \(-\frac{522744}{1024}\)
- Đề bài có sai sót gì không bạn ?
a) 125 x 25 x 64
= ( 25 x 64 ) x 125
= 1600 x 125
= 200 000
b) 125 x 25 x 12 x 14
= ( 125 x 12 ) x ( 25 x 14 )
= 1500 x 350
= 525000
c) 1300 : 50 = 26
d) 700 : 25 = 28
a, = 125 x (25x64)=125 x 1600= 200 000
b, = (125 x 24) x ( 25 x 12) = 3000 x 300 = 900 000
c, 1300 : 50 = 26
d, 700 : 25 = 28.
\(x^{64}=x\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}\)
vậy \(\hept{\begin{cases}x=0\\x=1\end{cases}}\)
\(\left(x-1\right)^4=9\left(x-1\right)^2\)
\(\left(x-1\right)^4-9\left(x-1\right)^2=0\)
\(\left(x-1\right)^2.\left[\left(x-1\right)^2-3^2\right]=0\)
\(\left(x-1\right)^2.\left(x-1-3\right)\left(x-1+3\right)=0\)
\(\left(x-1\right)^2.\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left(x-1\right)^2=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}}\)
\(\Rightarrow x-1=0\)hoặc \(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
vậy \(x=1\)hoặc \(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
x = -1;0;1.
x = 0 .