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19 tháng 8 2018
Bài 4 : Tìm x biết:
a, 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7)(2x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b, x2 + 36 = 12x
\(\Leftrightarrow\) x2 + 36 - 12x = 0
\(\Leftrightarrow\) x2 - 2.x.6 + 62 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x = 6
19 tháng 8 2018
e, (x - 2)2 - 16 = 0
\(\Leftrightarrow\) (x - 2)2 - 42 = 0
\(\Leftrightarrow\) (x - 2 - 4)(x - 2 + 4) = 0
\(\Leftrightarrow\) (x - 6)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
f, x2 - 5x -14 = 0
\(\Leftrightarrow\) x2 + 2x - 7x -14 = 0
\(\Leftrightarrow\) x(x + 2) - 7(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(x - 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)
3 tháng 1 2023
1: =>x^3-5x^2+x^2-5x+3x-15=0
=>(x-5)(x^2+x+3)=0
=>x-5=0
=>x=5
2: =>x^3+6x^2+12x+35=0
=>x^3+5x^2+x^2+5x+7x+35=0
=>(x+5)(x^2+x+7)=0
=>x+5=0
=>x=-5
3: \(\Leftrightarrow\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}+1\right)\)
=>x+100=0
=>x=-100
LA
4 tháng 2 2020
a) Đặt x -3 = a
<=> a(a+2)(a+8)(a+10) - 297=0
<=> \(\left[a\left(a+10\right)\right]\left[\left(a+2\right)\left(a+8\right)\right]\)-297=0
<=> \(\left(a^2+10a\right)\left(a^2+10a+16\right)-297=0\)
Đặt \(a^2+10a=b\)
\(b^2+16b-297=0\)
\(\Rightarrow\left[{}\begin{matrix}b=11\\b=-27\end{matrix}\right.\)\(b=11\Rightarrow\left[{}\begin{matrix}a=1\\a=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b= -27 \(\Rightarrow a=\varnothing\Rightarrow x=\varnothing\)
LA
4 tháng 2 2020
b) bấm máy ra nhân tử chung :D
c)
\(\Leftrightarrow\left(\frac{1927-X}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+...=0\)
\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
<=> x = 2018
d) \(\Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-3\right)=0\)
giống câu c
15 tháng 7 2018
a) \(4.\left(x-1\right)^2-9=0\)
\(\Rightarrow4.\left(x-1\right)^2=9\)
\(\Rightarrow\left(x-1\right)^2=9:4=\dfrac{9}{4}=\left(\pm\dfrac{3}{2}\right)^2\)
\(\Rightarrow x-1=\pm\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) \(\dfrac{1}{4}-9.\left(x-1\right)^2=0\)
\(\Rightarrow9.\left(x-1\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left(x-1^2\right)=\dfrac{1}{36}=(\pm\dfrac{1}{6})^2\)
\(\Rightarrow x-1=\pm\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{6}\\x-1=\dfrac{-1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
15 tháng 7 2018
e) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\)
\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2=\dfrac{1}{16}=\left(\pm\dfrac{1}{4}\right)^2\)
\(\Rightarrow2x+\dfrac{3}{4}=\pm\dfrac{1}{4}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}2x+\dfrac{3}{4}=\dfrac{1}{4}\\2x+\dfrac{3}{4}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
27 tháng 6 2018
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
20 tháng 6 2022
e:
Tham khảo: 
a: \(\Leftrightarrow x^2-2x+1+4x^2+4x+4-5x^2+5=0\)
\(\Leftrightarrow2x+10=0\)
hay x=-5
\(x^2-\dfrac{1}{49}=0\\ \Leftrightarrow x^2=\dfrac{1}{49}=\left(\dfrac{1}{7}\right)^2=\left(-\dfrac{1}{7}\right)^2\\\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{1}{7}\end{matrix}\right.\\ 64-0,25x^2\\ \Leftrightarrow0,25x^2=64\\ x^2=\dfrac{64}{0,25}=256=16^2=\left(-16\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)
\(x^2-\dfrac{1}{49}=0\\ \Rightarrow x^2=\dfrac{1}{49}\\ \Rightarrow x^2=\pm\dfrac{1}{7}\)
\(64-0,25x^2=0\\ \Rightarrow\left(0,5x\right)^2=64\\ \Rightarrow\left(0,5x\right)=\pm8\\ \Rightarrow x=\pm16\)