\(\left(3-4x\right)^2=25=5^2\)

\(\Rightarrow3-4x=5\)

\(\Rightarrow4x=3-5=-2\Rightarrow x=-\frac{1}{2}\)

\(\left(2x-\frac{1}{4}\right)^2=16=4^2\)

\(\Rightarrow2x-\frac{1}{4}=4\Rightarrow2x=4+\frac{1}{4}=\frac{17}{4}\)

\(\Rightarrow x=\frac{17}{4}:2=\frac{17}{4}.\frac{1}{2}=\frac{17}{8}\)

Đề số 3 bị sai.

\(\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=-\frac{5}{2}\)

(3-4x)²=25

3-4x=5

4x=3-5

4x=-2

x=-2:4

x=-0,5

b)(2x-1/4²)=16

2x-1/4=4

2x=4+1/4

2x=4,25

x=2,125

c) cái này x ở đâu vậy bn

d) (2x+5)²=0

2x+5=0

2x=0+5

2x=5

x=5:2

x=5/2

Nhớ k cho mk nha

Bài 1:

a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)

\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)

\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)

\(=-\frac{891}{25}.\frac{1}{4}\)

\(=-\frac{891}{100}\)

b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)

\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)

\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)

\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)

\(=\frac{3}{8}.\left(-14\right)\)

\(=-\frac{21}{4}\)

c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)

\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)

\(=1+1=2\)

d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)

\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)

\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)

\(=1+1=2\)

câu 2: 2x^3-1=5 

=>x=2

=>......=>.....

y=57; z=41

x+y+z=100

• Ngon thì tự đi mà giải

h) \(\left(x-1\right)^2=25\)

Mà:\(5^2=\left(-5\right)^2=25\)

TH1:\(x-1=5\)

\(x=5+1\)

\(x=6\)

TH2:\(x-1=-5\)

\(x=-5+1\)

\(x=-4\)

Vậy:\(x=6\)hoặc \(x=-4\)

i)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

Mà:\(\left(\frac{2}{5}\right)^2=\left(\frac{-2}{5}\right)^2=\frac{4}{25}\)

TH1:\(x+\frac{1}{2}=\frac{2}{5}\)                                       TH2:\(x+\frac{1}{2}=\frac{-2}{5}\)

\(x=\frac{2}{5}-\frac{1}{2}\)                                                        \(x=\frac{-2}{5}-\frac{1}{2}\)

\(x=\frac{-1}{10}\)                                                              \(x=\frac{-9}{10}\)

Vậy:\(x=\frac{-1}{10}\)hoặc\(x=\frac{-9}{10}\)

\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)

\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)

\(=\sqrt{\frac{13}{16}}:5\)

\(=\frac{\sqrt{13}}{4}:5\)

\(=\frac{\sqrt{13}}{20}\)

a, \(\left(2x-1\right)=-8\)

\(2x=-8+1\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

a) (2x - 1) = -8

⇒ 2x = -8 + 1

⇒ 2x = -7

b) (3x - 2)\(^2\) = \(\frac{1}{49}\)

Ta có: \(\frac{1}{49}\) = \(\frac{1}{7}\). \(\frac{1}{7}\) hoặc \(\frac{1}{49}\) = \(\frac{-1}{7}\). \(\frac{-1}{7}\)

TH1: 3x - 2 = \(\frac{1}{7}\) TH2: 3x - 2 = \(\frac{-1}{7}\)

⇒ 3x = \(\frac{1}{7}\)+2 ⇒ 3x = \(\frac{-1}{7}\)+2

⇒ 3x = \(\frac{15}{7}\) ⇒ 3x = \(\frac{13}{7}\)

⇒ x = \(\frac{5}{7}\) ⇒ x = \(\frac{13}{21}\)

Vậy: x = \(\frac{5}{7}\) hoặc x = \(\frac{13}{21}\)

Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}