√(√(81))-√(√(256))+√(12^2+5^2)
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a: =5*1,4-4*1,5+3*1,3
=7-6+3,9=4,9
b: =1/3*7+3/4*18-2/3*20
=7/3+54/4-40/3
=-11+54/4
=2,5
c: =5/6*17-1/2*16+2/5*15
=85/6-8+6
=85/6-2
=73/6
d: =15*4/5+12*3/4-18*4/9
=12+9-8
=12+1=13
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
256 ÷ 81 × 2 + 22 =50
256 x 2: 81 + 22 = 50
512 : 81 = 50 - 22
512/81 = 28
Vô lí
a: \(\Leftrightarrow2^3< 2^x< 2^4\)
=>3<x<4
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
=>12-x=4
hay x=8
c: \(\Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{2}{5}\right)^2=\left(\dfrac{2}{5}\right)^5\)
=>x>5
d: \(\Leftrightarrow3x-1=-4\)
=>3x=-3
hay x=-1
a: \(\Leftrightarrow2^3< 2^x< 2^4\)
=>3<x<4
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
=>12-x=4
hay x=8
c: \(\Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{2}{5}\right)^2=\left(\dfrac{2}{5}\right)^5\)
=>x>5
d: \(\Leftrightarrow3x-1=-4\)
=>3x=-3
hay x=-1
`= sqrt 9 - sqrt 16 + 13`.
`= 3-4+13`
`=12`.