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a,
\(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
b,
\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)
c,

bài này họ cho có hình không ạ? hay mình phải tự vẽ ạ?


Đặt B = 42004 + 42003 + 42002 + 42001 + ... + 42 + 4 + 1 (có 2005 số; 2005 chia 2 dư 1)
B = 42003.(4 + 1) + 42001.(4 + 1) + ... + 4.(4 + 1) + 1
B = 42003.5 + 42001.5 + ... + 4.5 + 1
B = 5.(42003 + 42001 + ... + 4) + 1
=> B = 5 x k + 1 (k thuộc N*; k chia hết cho 4)
=> A = 75 x (5 x k + 1) + 25
=> A = 75 x 5 x k + 75 + 25
=> A = (...00) + 100
=> A = (...00) chia hết cho 100
Có j thắc mắc thêm cứ hỏi

a,(0,8)5:(0,4)6 = (\(\dfrac{0,8}{0,4}\))5 : 0,4 = 25:0,4 = 80
b, (-25)7: 323 - \(\dfrac{6103515625}{3276}\) = - 186264,5149
c, \(\dfrac{4^2.4^3}{2^{10}}\) = \(\dfrac{4^5}{2^{10}}\) = \(\dfrac{2^{10}}{2^{10}}\) = 1
d, \(\dfrac{9^5.5^7}{45^7}\) = \(\dfrac{9^5.5^7}{9^7.5^7}\) = \(\dfrac{1}{81}\)
\(\dfrac{\left(0.8\right)^5}{\left(0.4\right)^4}\)=\(\dfrac{\left(2.0,4\right)^5}{\left(0.4^4\right)}\)=\(\dfrac{2^5.\left(0.4\right)^5}{\left(0,4\right)^4}\)=\(2^5\).\(\left(0.4\right)^1\)=12,8
b)câu b không biết có sai đề không nhưng đáp án câu b là -186264,5149
c) \(\dfrac{4^2.4^3}{2^{10}}\)=\(\dfrac{4^5}{\left(2^2\right)^5}\)=\(\dfrac{4^5}{4^5}\)=1
d)\(\dfrac{9^5.5^7}{45^7}\)=\(\dfrac{9^5.5^5.5^2}{45^7}\)=\(\dfrac{45^5.5^2}{45^7}\)=\(\dfrac{5^2}{45^2}\)=\(\left(\dfrac{5}{45}\right)^2\)=\(\left(\dfrac{1}{9}\right)^2\)=\(\dfrac{1}{81}\)

A = |\(x\) + 5| + 2023
|\(x\) + 5| ≥ 0 ⇒| \(x\) + 5| + 2023 ≥ 2023⇒ A(min) = 2023 xảy ra khi \(x\) = -5
B = (\(x+2\))2 - 2023
(\(x\) + 2)2 ≥ 0 ⇒ (\(x\) + 2)2 ≥ - 2023 ⇒ A(min) = -2023 xảy ra khi \(x\) = -2
C = \(x^2\) - 6\(x\) + 20
C = (\(x^2\) - 3\(x\)) - ( 3\(x\) - 9) + 11
C = \(x\)(\(x-3\)) - 3(\(x\) -3) + 11
C = (\(x-3\))(\(x\)-3) + 11
C = (\(x-3\))2 + 11
(\(x\) -3)2 ≥ 0 ⇒ (\(x\) - 3)2 + 11 ≥ 11 vậy C(min) = 11 xảy ra khi \(x=3\)
D = \(x^2\) + 10\(x\) - 25
D = \(x^2\) + 5\(x\) + 5\(x\) + 25 - 55
D = (\(x^2\) + 5\(x\)) + (5\(x\) + 25) - 50
D = \(x\)(\(x\) + 5) + 5(\(x\) + 5) - 50
D = (\(x\) +5)(\(x\) + 5) - 50
D = ( \(x\) + 5)2 - 50
(\(x+5\))2 ≥ 0 ⇒ (\(x\) + 5)2 - 50 ≥ -50 ⇒ D(min) = -50 xảy ra khi \(x\) = -5

\(A=\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)\left(\frac{1}{36}-1\right)...\left(\frac{1}{100}-1\right)\)
\(-A=\left(1-\frac{1}{16}\right)\left(1-\frac{1}{25}\right)\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{100}\right)\)
\(-A=\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\cdot...\cdot\frac{99}{100}\)
\(-A=\frac{\left(3\cdot5\right)\left(4\cdot6\right)\left(5\cdot7\right)...\left(9\cdot11\right)}{\left(4\cdot4\right)\left(5\cdot5\right)\left(6\cdot6\right)...\left(10\cdot10\right)}\)
\(-A=\frac{\left(3\cdot4\cdot5\cdot...\cdot9\right)\left(5\cdot6\cdot7\cdot...\cdot11\right)}{\left(4\cdot5\cdot6\cdot...\cdot10\right)\left(4\cdot5\cdot6\cdot...\cdot10\right)}\)
\(-A=\frac{3\cdot11}{10\cdot4}=\frac{33}{40}\)
\(A=-\frac{33}{40}\)