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29 tháng 8 2021
\(x^2-\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\\ \Rightarrow x^2-\dfrac{1}{4}=\dfrac{3}{4}\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=-\dfrac{1}{3}\\\dfrac{1}{2}-x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)
\(x^2-5=5\\ \Rightarrow x^2=10\\ \Rightarrow x=\pm\sqrt{10}\)
\(3x^2-1=14\\ \Rightarrow3x^2=15\\ \Rightarrow x^2=5\\ \Rightarrow x=\pm\sqrt{5}\)
KK
29 tháng 8 2021
1. x2 - \(\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\)
<=> x2 - \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\)
<=> x2 = \(\dfrac{3}{4}+\dfrac{1}{4}\)
<=> x2 = 1
<=> x = \(\pm\)1
2. \(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\)
<=> \(\dfrac{1}{4}-x+x^2=\dfrac{1}{9}\)
<=> x2 - x = \(\dfrac{1}{9}-\dfrac{1}{4}\)
<=> x2 - x = \(\dfrac{-5}{36}\)
<=> x2 - x - \(\dfrac{-5}{36}\) = 0
Đoạn này dài, mik giải ngoài rồi viết vào nha:
<=> x = \(\dfrac{5}{6}\)
3. x2 - 5 = 5
<=> x2 = 10
<=> x = \(\sqrt{10}\)
4. 3x2 - 1 = 14
<=> 3x2 = 15
<=> x2 = 15 : 3
<=> x2 = 5
<=> x = \(\sqrt{5}\)
VN
Võ Ngọc Phương
CTVHS
VIP
1 tháng 10 2023
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
1 tháng 10 2023
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
17 tháng 5 2022
a: \(A=2\cdot2^2-\dfrac{1}{3}\cdot9=8-3=5\)
b: \(B=\dfrac{1}{2}a^2-3b^2=\dfrac{1}{2}\cdot4-3\cdot\dfrac{1}{9}=2-\dfrac{1}{3}=\dfrac{5}{3}\)
5 tháng 4 2018
a/ Ta có: P(x)=0
nên 4x2 - 3x=0
do đó: 4xx-3x=0
(4x-3)x=0
Suy ra: 4x-3 = 0 hoặc x=0
=> x=\(\dfrac{3}{4}\) hoặc x=0
Vậy x=\(\dfrac{3}{4}\) hoặc x=0 là nghiệm của P(x)
b/ P(x)=0
2x2-8x=0
Nên (2x-8)x=0
=> 2x-8=0 hoặc x=0
Do đó: x=4 hoặc x=0
Vậy x=4 hoặc x=0 là nghiệm của P(x)
c/ P(x)=0
7x-2x2=0
(7-2x)x=0
Nên 7-2x=0 hoặc x = 0
Do đó: x=\(\dfrac{7}{2}\) hoặc x = 0
Vậy x=\(\dfrac{7}{2}\) hoặc x = 0 là nghiệm của P(x)
d/ Ta có: P(x)=0
nên \(\dfrac{3}{4}x-\dfrac{1}{2}x^2=0\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}x\right)x=0\)
Do đó: \(\dfrac{3}{4}-\dfrac{1}{2}x=0\) hoặc x=0
Suy ra: x= \(\dfrac{3}{2}\) hoặc x=0
Vậy x= \(\dfrac{3}{2}\) hoặc x=0 là nghiệm của P(x)
ND
Nguyễn Đức Trí
VIP
31 tháng 7 2023
\(4.3^x+3^{x+1}=63\)
\(\Rightarrow4.3^x+3.3^x=63\)
\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)
\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)
mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)
\(\Rightarrow x\in\varnothing\)
25 tháng 10 2021
\(\dfrac{3x-2}{4}=\dfrac{9}{3x-2}\\ \Rightarrow\left(3x-2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}3x-2=-6\\3x-2=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{8}{3}\end{matrix}\right.\)
25 tháng 10 2021
\(\dfrac{3x-2}{4}=\dfrac{9}{3x-2}\left(đk:x\ne\dfrac{2}{3}\right)\)
\(\Rightarrow\left(3x-2\right)^2=36\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=6\\3x-2=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
12 tháng 9 2021
a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)
b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)
c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)
d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
\(\left(3x-\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\\\left(3x-\dfrac{1}{2}\right)^2=\left(-\dfrac{2}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{2}=\dfrac{2}{3}\\3x-\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{7}{6}\\3x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{18}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
Vậy ..
thank nha Nguyễn Thanh Hằng tồ