nêu chi v

/

thi rồi nêu lên chi

(2\(x\) - 1)\(^2\) - (4\(x^2\) - 1) = 0

4\(x^2-4x+1\) - 4\(x^2+1\) = 0

(4\(x^2-4x^2\)) - 4\(x\) + (1+ 1) = 0

0 - 4\(x\) + 2 = 0

4\(x=2\)

\(x=2:4\)

\(x=\frac12\)

Vậy \(x=\frac12\)

3

chiều rộng khu rừng là:
4:2=2 ( km)

diện tích khu rừng HCN là:

4 x 2 = 8 (km2)

đáp số: 8( km2)

Chiều rộng khu rừng là:
4:2=2 ( km)

diện tích khu rừng hình chữ nhật là:

4 x 2 = 8 (km^2)

Đáp số: 8( km^2)

\(\frac{23}{1000}\) = 0,023

\(\frac{23}{1000}=0,023\)

2 + 3 x [(\(\frac23\) - \(\frac12\)) : \(\frac76\) + \(\frac{13}{7}\)]

= 2 + 3 x [(\(\frac46-\frac36\)) x \(\frac67\) + \(\frac{13}{7}\)]

= 2+ 3 x [\(\frac16\times\frac67\) + \(\frac{13}{7}\)]

= 2 + 3 x [\(\frac17+\frac{13}{7}\)]

= 2 + 3 x 2

= 2 + 6

= 8

Iavva

1258 x 5+ 1258 x 4 + 1258

= 1258 x 5 + 1258 x 4 + 1258 x 1

= 1258 x (5 + 4 + 1)

= 1258 x (9 + 1)

= 1258 x 10

= 12580

31450

Câu 18:

{109 + 14765 + [2.(35 - 8+ 13)]}

= {109 + 14765 + [2.(27 + 13)]}

= {109 + 14765 + [2.40]}

= {109 + 14765 + 80}

= 14784 + 80

= 14954

50 x 800 - 123.[678.123]

= 40 000 - 123.83394

= 40 000 - 102574462

= - 10217462

-10 217 462✔

☺✔

Tìm \(x\) biết: (\(2x-4\))\(^2\) = 642050\(^0\)

(2\(x\) - 4)\(^2\) = 1

\(\left[\begin{array}{l}2x-4=-1\\ 2x-4=1\end{array}\right.\)

\(\left[\begin{array}{l}2x=-1+4\\ 2x=1+4\end{array}\right.\)

\(\left[\begin{array}{l}2x=3\\ 2x=5\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac32\\ x=\frac52\end{array}\right.\)

Vậy \(x\in\) {\(\frac32;\frac52\)}

(2x-4)^2=642025^0

(2x-4)^2=1

(2x-4)^2=1^2

2x-4=1

2x=1+4

2x=5

x=5/2

Tìm \(x\) biết:

(\(x+\frac12\))\(^2\) = \(\frac{1}{16}\)

(\(x+\frac12\))\(^2\) = \(\left(\frac14\right)^2\)

\(\left[\begin{array}{l}x+\frac12=-\frac14\\ x+\frac12=\frac14\end{array}\right.\)

\(\left[\begin{array}{l}x=-\frac14-\frac12\\ x=\frac14-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=-\frac14-\frac24\\ x=\frac14-\frac24\end{array}\right.\)

\(\left[\begin{array}{l}x=-\frac34\\ x=-\frac14\end{array}\right.\)

Vậy \(x\) \(\in\) {- \(\frac34;-\frac14\)}

Ta có: \(\left(x+\frac12\right)^2=\frac{1}{16}\)

=>\(\left[\begin{array}{l}x+\frac12=\frac14\\ x+\frac12=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac14-\frac12=-\frac14\\ x=-\frac14-\frac12=-\frac34\end{array}\right.\)