Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\
\left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\) TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\) TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\
x=-\frac{3}{5}-\frac{1}{5}\\
x=-\frac{4}{5}\) \(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\) \(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\) \(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\) \(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\) \(\Rightarrow x=\frac{2}{5}\) \(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\) \(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\) \(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\) \(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\) \(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\) \(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\) \(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\) \(\Rightarrow3x=\frac{1}{9}\) \(\Rightarrow x=\frac{1}{27}\) \(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\) a)\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\) =\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\frac{3^2}{5^2}\) =\(x+\frac{1}{5}=\frac{3}{5}\) \(x=\frac{2}{5}\) b)\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{24}{27}\) =\(x=-\frac{35}{27}\) 3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\) => \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\) => \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\) => \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\) => \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\) => \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\) => \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\) => \(x\) = \(\dfrac{2}{5}\) 4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\) => \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\) => \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\) => \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\) => \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) => \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\) => \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\) => \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\) => \(3x=\dfrac{1}{9}\) => \(x=\dfrac{1}{9}:3\) => \(x=\dfrac{1}{27}\) a, <=>(-5)x=(-5)3 <=> x=3 b, <=> 52x=522 <=> x=11 c, 32x=317.315 <=> 32x=332 <=> x=16 d,2x+1=225 <=> x+1=25 <=> x=24 Chúc hok tốt!!! a,(5/8/17+-4/17):x+33/182=4/11 =5/4/17:x+33/182=4/11 5/4/17:x=4/11-33/182 5/4/17:x=365/2002 x=5/4/17:365/2002 x=28/4438/6205 b,-1/5/27-(3x-7/9)^3=-24/27 (3x-7/9)^3=-1/5/27--24/27 (3x-7/9)^3=-8/27 (3x-7/9)^3=(-2/3)^3 3x-7/9=-2/3 3x=-2/3+7/9 3x=1/9 x=1/9:3 x=1/27 Bài 2: a) \(\left(x-3\right)^3+27=0\) \(\Leftrightarrow\left(x-3\right)^3=0-27\) \(\Leftrightarrow\left(x-3\right)^3=-27\) \(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\) \(\Leftrightarrow x-3=-3\) \(\Leftrightarrow x=\left(-3\right)+3\) \(\Leftrightarrow x=0\) b) \(-125-\left(x+1\right)^3=0\) \(\Leftrightarrow\left(x+1\right)^3=-125-0\) \(\Leftrightarrow\left(x+1\right)^3=-125\) \(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\) \(\Leftrightarrow x+1=-5\) \(\Leftrightarrow x=\left(-5\right)-1\) \(\Leftrightarrow x=-6\) c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\) \(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\) \(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\) \(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\) \(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\) \(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\) \(\Leftrightarrow2x=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{1}{2}:2\) \(\Leftrightarrow x=\dfrac{1}{4}\) d) \(2^x+2^{x+1}=24\) \(\Leftrightarrow2^x+2^x.2=24\) \(\Leftrightarrow2^x\left(1+2\right)=24\) \(\Leftrightarrow2^x.3=24\) \(\Leftrightarrow2^x=24:3\) \(\Leftrightarrow2^x=8\) \(\Leftrightarrow2^x=2^3\) \(\Rightarrow x=3\) e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\) \(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\) \(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\) \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\) g) \(\left|x-3\right|+2x=10\) \(\Leftrightarrow\left|x-3\right|=10-2x\) \(\Leftrightarrow\left|x-3\right|=2.5-2x\) \(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\) (không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...) Bài 1: a) \(2^7+2^9⋮10\) Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\) \(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\) \(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\) \(\Leftrightarrow\overline{C8}+\overline{D2}\) \(\Leftrightarrow\overline{E0}\) Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\) b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\) Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\) \(\Leftrightarrow2^{72}.5^{20}\) Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\) c) \(3^{10}+3^{12}⋮30\) Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\) \(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\) \(\Leftrightarrow\overline{A1}.9+\overline{B1}\) \(\Leftrightarrow\overline{C9}+\overline{B1}\) \(\Leftrightarrow\overline{D0}⋮10\) (Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
K
Khách
CM
Cao Minh Tâm
15 tháng 8 2019
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