bn lấy máy tính mà tính ý

Bài1:

Ta có:

a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)

b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)

c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)

Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)

Bài 2:

Không có đề bài à bạn?

Bài 3:

a)\(\sqrt{x}-1=4\)

\(\Rightarrow\sqrt{x}=5\)

\(\Rightarrow x=\sqrt{25}\)

\(\Rightarrow x=5\)

b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)

Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)

\(\Rightarrow\left(x-1\right)^2=16\)

\(\Rightarrow\left(x-1\right)^2=4^2\)

\(\Rightarrow x-1=4\)

\(\Rightarrow x=5\)

bài 1 :
b) (x-1/2 )² = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) ² = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )³ = -8
<=> ( 2x - 1) ³ = ( -2 ) ³
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2

Bài 2 :
c) 3^2x-1=243
<=> 3^2x-1= 3⁵
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3

Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !

mấy câu kia cần nữa k

bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)

\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)

vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)

bài 1

\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)

\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)

\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)

\(\left(-2\right)\cdot\left\lbrack\frac{x}{8}-\left(-\frac13\right)\right\rbrack+\frac32=\frac14\)

\(\left(-2\right)\cdot\left(\frac{x}{8}+\frac13\right)+\frac32=\frac14\)

\(\left(-2\right)\cdot\left(\frac{x}{8}+\frac13\right)=\frac14-\frac32\)

\(\left(-2\right)\cdot\left(\frac{x}{8}+\frac13\right)=-\frac54\)

\(\left(\frac{x}{8}+\frac13\right)=-\frac54:\left(-2\right)\)

\(\left(\frac{x}{8}+\frac13\right)=-\frac54\cdot\left(-\frac12\right)\)

\(\left(\frac{x}{8}+\frac13\right)=\frac58\)

\(\frac{x}{8}=\frac58-\frac13\)

\(\frac{x}{8}=\frac{7}{24}\)

\(x=\frac{7\cdot8}{24}=\frac73\)

vậy \(x=\frac73\)

(-2).(x/8-(-1/3) + 3/2 =1/4

(-2).(x/8-(-1/3) = 1/4-3/2

(-2).(x/8-(-1/3) = 1/4 - 6/4

(-2).(x/8-(-1/3) = -5/4

x/8-(-1/3) = -5/4 ; (-2)

x/8-(-1/3) = 5/8

x/8 = 5/8 - 1/3

x/8 =15/24- 8/24 =7/24

=> 3x/24 = 7/24

=> 3x=7

x= 7/3

vậy x = 7/3

Bài 1:

a) \(x^2-3=1\)

\(\Rightarrow x^2=1+3=4\)

\(\Rightarrow x=\pm2\)

b)\(2x^3+12=-4\)

\(\Rightarrow2x^3=-4-12=-16\)

\(\Rightarrow x^3=-8\)

\(\Rightarrow x=-2\)

c)\(\left(2x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)

b) \(2x^3+12=-4\Rightarrow2x^3=-16\)

\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)

\(\Rightarrow x=-2\)

c) \(\left(2x-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d,h,i,k cững tương tự....

a) \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)

=> \(\left(7-8x\right).5=6\left(-4+2x\right)\)

=> 35 - 40x = -24 + 12x

=> 35 + 24 = 12x + 40x

=> 52x = 59

=> x = 59/52

b) \(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)

=> (1 - 3: x)² = 8²

=> \(\orbr{\begin{cases}1-3:x=8\\1-3:x=-8\end{cases}}\)

=> \(\orbr{\begin{cases}3:x=-7\\3:x=9\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{7}\\x=\frac{1}{3}\end{cases}}\)

c) \(\left(x+1\right)\left(x-2\right)\ge0\)

=> \(\hept{\begin{cases}x+1\ge0\\x-2\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-2\le0\end{cases}}\)

=> \(\hept{\begin{cases}x\ge-1\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\le2\end{cases}}\)

=> \(-1\le x\le2\)

h) \(\left(x+1\right)\left(x-3\right)\le0\)

=> \(\hept{\begin{cases}x+1\ge0\\x-3\le0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-3\ge0\end{cases}}\)

=> \(\hept{\begin{cases}x\ge-1\\x\le3\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}\) (loại)

= \(-1\le x\le3\)

(-2).(x/8-(-1/3) + 3/2 =1/4

(-2).(x/8-(-1/3) = 1/4-3/2

(-2).(x/8-(-1/3) = 1/4 - 6/4

(-2).(x/8-(-1/3) = -5/4

x/8-(-1/3) = -5/4 ; (-2)

x/8-(-1/3) = 5/8

x/8 = 5/8 - 1/3

x/8 =15/24- 8/24 =7/24

=> 3x/24 = 7/24

=> 3x=7

x= 7/3

vậy x = 7/3

Ta có: \(\left(-2\right)\left(\frac{x}{8}-\frac{-1}{3}\right)+\frac32=\frac14\)

=>\(-2\left(\frac{x}{8}+\frac13\right)=\frac14-\frac32=\frac14-\frac64=-\frac54\)

=>\(\frac{x}{8}+\frac13=\frac58\)

=>\(\frac{x}{8}=\frac58-\frac13=\frac{15}{24}-\frac{8}{24}=\frac{7}{24}\)

=>\(x=\frac{7}{24}\cdot8=\frac73\)

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