mn ơi tính hộ mk bài tìm x với mk đang cần gấp::2 x 3X + 1=10 x 312 + 8 x 312
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\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
\(\left(3x-1\right)^2-9^2=13\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\sqrt{94}\\3x-1=-\sqrt{94}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{94}+1}{3}\\x=\dfrac{-\sqrt{94}+1}{3}\end{matrix}\right.\)
\(2^x:1+2^x:2+...+2^x:49=2^{49}-1\)
\(2^x.1+2^x.\frac{1}{2}+...+2^x.\frac{1}{49}=2^{49}-1\)
\(2^x.\left(1+\frac{1}{2}+...+\frac{1}{49}\right)=2^{49}-1\)
Đặt: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{49}}\right)\)
=> \(A=1-\frac{1}{2^{49}}=\frac{2^{49}-1}{2^{49}}\)
\(2^{x-1}+2^{x-2}+2^{x-3}+...+2^{x-49}=2^{49}-1\)
<=> \(\frac{2^x}{2}+\frac{2^x}{2^2}+\frac{2^x}{2^3}+...+\frac{2^x}{2^{49}}=2^{49}-1\)
<=> \(2^x\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\right)=2^{49}-1\)
<=> \(2^x.\frac{2^{49}-1}{2^{49}}=2^{49}-1\)
<=> \(2^x=2^{49}\)
<=> x = 49.
a)(x - 1) x + 2 = (x - 1)x + 4
=> (x - 1) x + 4 - (x - 1)x + 2 = 0
=> (x - 1)x + 2 . [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0^{x+2}\\\left(x-1\right)^2=1^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 0
=> x = 1
Nếu x - 1 = - 1
=> x = 0
Nếu x - 1 = 1
=> x = 2
Vậy \(x\in\left\{0;1;2\right\}\)
b) \(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\)
\(\Rightarrow1,78^{2x-2}:1,78^x-1,78^x:1,78^x=0\)
\(\Rightarrow1,78^{x-2}-1=0\)
\(\Rightarrow1,78^{x-2}=1\)
\(\Rightarrow1,78^{x-2}=1,78^0\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy x = 2
\(2.3^{x+1}=10.3^{12}+8.3^{12}\\ \Rightarrow2.3^{x+1}=3^{12}\left(10+8\right)\\ \Rightarrow2.3^{x+1}=3^{12}.18\\ \Rightarrow3^{x+1}=9.3^{12}\\ \Rightarrow3^{x+1}=3^{14}\\ \Rightarrow x+1=14\\ \Rightarrow x=13\)