\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)

\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)

\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)

\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)

\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)

\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)

g) \(\sqrt[]{64}+2\sqrt[]{\left(-3\right)^2}-7\sqrt[]{1,69}+3\sqrt[]{\dfrac{25}{16}}\)

\(=8+2.3-7.1,3+3.\dfrac{5}{4}\)

\(=14-9,1+\dfrac{15}{4}\)

\(=5,1+3,75=8,85\)

ko tính đc

số âm ko có căn bậc 2 số học

\(\sqrt[]{2^2+\sqrt[]{4^2}+\sqrt[]{\left(-6\right)^2}+\sqrt[]{\left(-8\right)^2}}\)

\(=\sqrt[]{4+4+6+8}=\sqrt[]{22}\)

\(a=2022.\left|x^2+1\right|+2023\)

\(\Rightarrow a=2022.\left(x^2+1\right)+2023\left(\left|x^2+1\right|>0,\forall x\right)\)

mà \(\left(x^2+1\right)\ge1,\forall x\)

\(\Rightarrow a=2022.\left(x^2+1\right)+2023\ge2022.1+2023=4045\)

\(\Rightarrow GTNN\left(a\right)=4045\left(x=0\right)\)

GTNN(a) = 4045 khi x = 0

a) \(x:\dfrac{1}{2}=\left(-\dfrac{1}{2}\right)^4\Rightarrow x:\dfrac{1}{2}=\dfrac{1}{16}\Rightarrow x=\dfrac{1}{16}.2=\dfrac{1}{8}\)

b) \(\left(-\dfrac{4}{7}\right)^5.x=\left(\dfrac{4}{7}\right)^7\Rightarrow-\left(\dfrac{4}{7}\right)^5.x=\left(\dfrac{4}{7}\right)^7\Rightarrow x=-\left(\dfrac{4}{7}\right)^7:\left(\dfrac{4}{7}\right)^5\Rightarrow x=-\left(\dfrac{4}{7}\right)^2=-\dfrac{16}{49}\)

Đính chính câu a

\(x:\dfrac{1}{2}=\dfrac{1}{16}\Rightarrow x=\dfrac{1}{16}.\dfrac{1}{2}=\dfrac{1}{32}\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

để A = 3x + 2/x - 3 nguyên

=>  3x + 2 ⋮ x - 3

=> 3x - 9 + 11 ⋮ x - 3

=> 3(x - 3) + 11 ⋮ x - 3

=> 11 ⋮ x - 3

=> x - 3 thuộc Ư(11)

=> x - 3 thuộc {-1; 1; -11; 11}

=> x  thuộc {2; 4; -8; 14}

Ftea.me am làm đúng rồi. cô tick xanh cho em nhưng lần sau em nhớ thêm đkxđ : \(x\ne\) 3

\(\dfrac{36}{35}=1,0\left(285714\right)\)

\(\dfrac{10}{15}=\dfrac{2}{3}=0,\left(6\right)\)

\(\dfrac{5}{11}=0,\left(45\right)\)

\(\dfrac{2}{13}=0,\left(153846\right)\)

\(\dfrac{15}{82}=0,1\left(82926\right)\)

\(\dfrac{13}{22}=0,5\left(90\right)\)

\(\dfrac{1}{60}=0,01\left(6\right)\)

\(\dfrac{5}{24}=0,208\left(3\right)\)

Bài 3A:

\(a,\left(x-1,2\right)^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x-1,2=2\\x-1,2=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3,2\\x=-0,8\end{matrix}\right.\\ b,\left(x+1\right)^3=-125\\ \Leftrightarrow x+1=-5\\ \Leftrightarrow x=-6\\ c,3^{4-x}=27\\ \Leftrightarrow4-x=3\\ \Leftrightarrow x=1\)

Bài 3A:

\(d,\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\\ e,3^{-1}\cdot4^x=\dfrac{5}{3}\cdot2^7\\ \Leftrightarrow4^x=2^7\cdot5\)

Em xem lại đề bài câu này không có x nguyên được.

\(f,9^{-x}\cdot27^x=27\\ \Leftrightarrow3^x=27\\ \Leftrightarrow x=3\)