nhờ mọi người tiếp ạ , và hãy trả lời cụ thể nha .
1) tìm x bt
a, (x + 5).(x - 4) = 0 b, (3 - x) . (x - 3) = 0 c, 5. ( x + 1) - 3mũ2 = 3 . (2 + x) - 8
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Ta có: \(3-2\left(x+1\right)=51\)
\(\Rightarrow2\left(x+1\right)=3-51\)
\(\Rightarrow2\left(x+1\right)=-48\)
\(\Rightarrow x+1=-48:2\)
\(\Rightarrow x+1=-24\)
\(\Rightarrow x=-24-1\)
\(\Rightarrow x=-25\)
Vậy \(x=-25.\)
a) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c)\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
d) \(x^3+x=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
e)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
Bài 1 :
A ) 3 < x < 5
=> x thuộc { 4 }
Vậy x = 4
Câu b và câu c cứ theo vậy mà làm .
Bài 2 :
| x + 7 | = 0
x = 0 - 7
x = -7
Vậy x = -7
a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
Bài 1:
\(a,\left(a+b-c\right)-\left(b-c-d\right)\)
\(=a+b-c-b+c+d\)
\(=a+d\)
\(b,-\left(a-b+c\right)+\left(a-b+d\right)\)
\(=-a+b-c+a-b+d\)
\(=-c+d\)
\(c,\left(a+b\right)-\left(-a+b-c\right)\)
\(=a+b+a-b+c\)
\(=2a+c\)
\(d,-\left(a+b\right)+\left(a+b+c\right)\)
\(=-a-b+a+b+c\)
\(=c\)
Bài 3 :
\(a,15-\left(4-x\right)=6\)
\(4-x=15-6\)
\(4-x=9\)
\(x=4-9\)
\(x=-5\)
\(b,-30+\left(25-x\right)=-1\)
\(25-x=-1+30\)
\(25-x=29\)
\(x=25-29\)
\(x=-4\)
\(c,x-5=-1\)
\(x=-1+5\)
\(x=4\)
\(d,x-4=-10\)
\(x=-10+4\)
\(x=-6\)
\(e,x+3=-8\)
\(x=-8-3\)
\(x=-11\)
\(g,x+6=0\)
\(x=-6\)
Câu 1:
A, (a+b-c)-(b-c-d)
= a+b-c-b+c+d
= a+(b-b)+(c-c)
= a
B, -(a-b+c)+(a-b+d)
= -a+b-c+a+b+d
= (a-a)+(b+b)+d-c
= 2b+d-c
C, (a+b)-(-a+b-c)
= a+b+a-b+c
= (a+a)+(b-b)+c
= 2a+c
D, -(a+b)+(a+b+c)
= -a-b+a+b+c
= (-a+a)+(b-b)+c
= c
=
a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)
b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)
\(\Rightarrow-\frac{7}{10}x=-1\)
\(\Rightarrow x=\frac{10}{7}\)
c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)
a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0
Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0
Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5
x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)
x = 14/3 hoặc x = -3
b, 1/10 .x - 4/5 .x + 1 =0
x . (1/10 - 4/5) + 1 = 0
x . (-7/10) + 1 = 0
x . -7/10 =0 +1 = 1
x = 1 : (-7/10)
x = -10/7
c, (2x - 1/3 ) . (5x +2/7) = 0
Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0
Vậy : 2x = 1/3 hoặc 5x = 2/7
x = 1/3 : 2 hoặc x = 2/7 : 5
x = 1/6 hoặc x = 2/35
1, => x + 12 = 0 => x = -12
x - 3 = 0 => x = 3
=> x \(\in\) { -12; 3 }
1; (\(x\) + 12)(\(x\) - 3) = 0
\(\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -12; 3}
a) Ta có: \(\left(x+5\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\).
b) \(\left(3-x\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3-x=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow x=3\)
Vậy \(x=3\).
c) \(5\left(x+1\right)-3^2=3\left(2+x\right)-8\)
\(\Rightarrow5x+5-9=6+3x-8\)
\(\Rightarrow5x-3x=-5+9-8\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy \(x=-2.\)
giúp mk câu này nha
x . ( x + 1 ) = 0