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14 tháng 7 2016
(x - 3)4 =16
=> 24 = 16
=> x- 3 = 2
=> x = 5
4(x-3) =16
=> 42 = 16
=> x-3 = 2
=> x = 5
KC
2
31 tháng 8 2023
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
ND
Nguyễn Đức Trí
VIP
31 tháng 8 2023
c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
8 tháng 7 2021
Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x^2+4x+16\right)=49\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+64=49\)
\(\Leftrightarrow-9x^2+27x-12=0\)
\(\Leftrightarrow-3\left(x^2-9x+4\right)=0\)
\(\Delta=\left(-9\right)^2-4\cdot1\cdot4=81-16=65\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{9-\sqrt{65}}{2}\\x_2=\dfrac{9+\sqrt{65}}{2}\end{matrix}\right.\)
7 tháng 7 2016
1. \(\frac{-17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
\(-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{7}{12}\)
\(-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)
\(-\frac{80}{84}< \frac{84x+48}{84}< \frac{49}{84}\)
\(-80< 84x+48< 49\)
\(\begin{cases}-80< 84x+48\\84x+48< 49\end{cases}\)
\(\begin{cases}84x>-128\\84x< 1\end{cases}\)
\(\begin{cases}x>-\frac{32}{21}\\x< \frac{1}{84}\end{cases}\)
\(\Rightarrow-\frac{32}{21}< x< \frac{1}{84}\)
PA
7 tháng 7 2016
\(-\frac{17}{21}\div\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
\(-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)
\(-\frac{32}{21}< x< \frac{1}{84}\)
\(-1^{11}_{21}< x< \frac{1}{84}\)
\(\Rightarrow x\in\left\{-1;0\right\}\)
Vậy x = 0
\(\frac{4}{3}\times1,25\times\left(\frac{16}{5}-\frac{5}{16}\right)< 2x< 4-\frac{4}{3}+3-\frac{3}{2}+2\)
\(\frac{77}{16}< 2x< \frac{37}{6}\)
\(\frac{77}{32}< x< \frac{37}{12}\)
\(2^{13}_{32}< x< 3^1_{12}\)
=> x = 3
PT
1
2 tháng 9 2016
Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)
DP
5 tháng 7 2017
\(x+\frac{1}{19}+x+\frac{2}{18}+x+\frac{3}{17}+x+\frac{4}{16}=4\)
\(4x+\left(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+\frac{4}{16}\right)=4\)
\(4x+\frac{6863}{11628}=4\)
\(4x=4-\frac{6863}{11628}=\frac{39649}{11628}\)
\(\Rightarrow x=\frac{39649}{11628\cdot4}=\frac{39649}{46512}\)
\(\frac{x^3}{16}=4\Rightarrow x^3=4.16\)
\(\Rightarrow x^3=64=4^3\)
=> x = 4
\(\frac{x^3}{16}=4=>x^3:16=4\)
\(x^3=64< =>4^3=64=>x=3\)