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a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)
\(5x+1=\pm\dfrac{6}{9}\)
+) \(5x+1=\dfrac{6}{9}\)
\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{9}\)
\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)
+) \(5x+1=\dfrac{-6}{9}\)
\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)
vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)
a: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
b: =>x-2/9=4/9
=>x=6/9=2/3
c: =>8x+1=5
=>8x=4
hay x=1/2
a,
- Theo đề bài ta có:
(8x-1)2n-1 = 52n-1
=> 8x-1 = 5
8x = 6
x = \(\dfrac{6}{8}\)= \(\dfrac{3}{4}\)
- Vậy x = \(\dfrac{3}{4}\)
b,
- Ta có:
(x - 7)x+1 - (x - 7)x+11 = 0
(x - 7)x . (x - 7) - (x - 7)x . (x - 7)11 = 0
(x - 7)x . [(x - 7) - (x - 7)11] = 0
=> (x - 7)x = 0 hoặc [(x - 7) - (x - 7)11] = 0
- TH1: (x - 7)x = 0
=> x - 7 = 0
=> x = 7
- TH2:
[(x - 7) - (x - 7)11] = 0
=> x - 7 = (x -7)11
=> x - 7 = 1 hoặc x - 7 = 0
+ Nếu x - 7 = 1
x = 8
+ Nếu x - 7 = 0 (TH1)
- Vậy x = 7 hoặc x = 8
c, - Theo đề bài ta có:
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
- Thấy \(\left(\dfrac{2}{3}\right)^6=\left(\dfrac{2}{3}\right)^{2\cdot3}\)= \(\left(\dfrac{4}{9}\right)^3\)
=> \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
=> \(x-\dfrac{2}{9}=\dfrac{4}{9}\)
=> \(x=\dfrac{4}{9}-\dfrac{2}{9}\)
\(x=\dfrac{2}{9}\)
- Vậy \(x=\dfrac{2}{9}\)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)
\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)
b) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow\left(x-\dfrac{2}{9}\right)^3=\left[\left(\dfrac{2}{3}\right)^2\right]^3=\left(\dfrac{4}{9}\right)^3\)
\(\Rightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{2}{3}\)
\(\text{làm hộ mik câu a bạn nha}\)