giúp mình với ^^
Tìm x:
a) 43x =39.43-37.43
b)3x-136=23.32+22.5
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\(5x\left(x-3\right)=x-3\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)
ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{x}.\)
\(A=1+\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{x}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2.2}+...+\frac{1}{x:2}\)
\(\Rightarrow2A-A=2-\frac{1}{x}\)
\(A=2-\frac{1}{x}=\frac{4095}{2048}\)
=> 1/x = 1/2048
=> x = 2048 ( 2048 = 211 )
a: \(x\left(1-2x\right)+2x^2=14\)
=>\(x-2x^2+2x^2=14\)
=>x=14
b: \(x\left(x-5\right)+3x-15=0\)
=>\(\left(x-5\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
\(2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{2}{x}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{2}{x}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{2}{x}+\frac{1}{x}\right)\)
=> \(A=2-\frac{1}{x}\)
Giải phương trình:
\(2-\frac{1}{x}=\frac{4095}{2048}\)
\(\frac{1}{x}=2-\frac{4095}{2048}\)
\(\frac{1}{x}=\frac{1}{2048}\)
x=2048
a)\(2x\left(x-2016\right)-2x+4032=0\)
\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)
\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)
b)\(5x\left(x-3\right)=x-3\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)
c)\(\left(3x-1\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x-1\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-1+x+2\right)\left[\left(3x-1\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
120 - 4x(y - 3) = 136
=> 4x(y - 3) = 136 - 120
=> 4x(y - 3) = 16
=> x(y - 3) = 16 : 4
=> x(y - 3) = 4
Mà 4 = 1.4 = 2.2 = (-1).(-4) = (-2).(-2)
Lập bảng xét tất cả 6 trường hợp :
\(x\) | \(1\) | \(4\) | \(2\) | \(-2\) | \(-1\) | \(-4\) |
\(y-3\) | \(4\) | \(1\) | \(2\) | \(-2\) | \(-4\) | \(-1\) |
\(x\) | \(1\) | \(4\) | \(2\) | \(-2\) | \(-1\) | \(-4\) |
\(y\) | \(7\) | \(4\) | \(5\) | \(1\) | \(-1\) | \(2\) |
Vậy các cặp (x,y) thỏa mãn là : (1;7) ; (2;5) ; (4;4) ; (-2 ; 1) ; (-1 ; -1) ; (- 4;2)
120 - 4(y - 3) = 136
<=> -4(y - 3) = 136 - 120
<=> -4(y - 3) = 16
<=> y - 3 = 16 : (-4)
<=> y - 3 = -4
<=> y = -4 + 3
<=> y = -1
=> y = -1
a) Ta có
43x=39.43-37.43
43x=43.(39-37)
43x=43x2
=>x=2
b) ta có
3x-136=2^3.3^2+2^2.5
3x-136=8.9+4.5
3x-136=72+20
3x-136=92
3x=228
x=76
a) 43x = 39.43 - 37.43
=> 43x = 43 . (39 - 37)
=> 43x = 43 . 2
=> x = 43 : 43 . 2
=> x = 1 . 2
Vậy x = 2