Tìm x a)(5x-3)(3x+1)-(15x+1)(x-2)=0 b)x²+(x+5)(x-3)-25=0
Những câu hỏi liên quan
Tìm x
a, (x+2)^3 - x(x^2+6x-3)=0
b,(x+4)^3 - x(x+6)^2=7
c, (x-1)^3 - x(x^2-3x-2)=0
d,1000x^3+90x(10x+3)+27=0
e,125x^3+15x(5x+1)= -1
f, 27x^3+45x(3x+5)+133=0
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
Tìm x biết a) x(x-25)=0 b)2x(x-4)-x(2x-1)=-28 c)x^2 -5x=0 d)(x-2)^2-(x+1)(x+3)=-7 e)(3x+5).(4-3x)=0 f)x^2-1/4=0
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
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bài 49; tìm x;
1, 3x ( x - 7) 2x - 14 = 0
2, x mũ 3 + 3x mũ 2 - ( x + 3) = 0
3, 15x - 5 + 6x mũ 2 - 2x = 0
4, 5x - 2 - 25x mũ 2 + 10x = 0
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
1) TÌm x biết:
a) ( 1/7x - 2/7).(-1/5x + 3/5).(1/3x + 4/3) =0
b) 1/6x + 1/10x -4/15x +1 =0
b, \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(x.0=-1\)
\(\Rightarrow x\in rỗng\)
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bài 9; tìm x
1, 3x( x - 7) 2x - 14 = 0
2, x mũ 3 + 3x mũ 2 - ( x + 3 )= 0
3, 15x - 5 + 6x mũ 2 - 2x =0
4, 5x - 2 - 25x mũ 2 + 10x = 0
a)1/6x + 1/10x - 4/15x + 1=0
b)(1/7x - 2/7)(-1/5x + 3/5)(1/3x + 4/3)=0
c)(x - 1/3)(y - 1/2)(z - 5)=0 và x + 2=y + 1=z + 3
a)\(\dfrac{1}{6}x+\dfrac{1}{10}x-\dfrac{4}{15}x+1=0\)
\(\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right).x+1=0\)
\(\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{8}{30}\right).x+1=0\)
\(0.x+1=0\)
\(0.x=-1\)
=> Không có giá trị nào của x.
Vậy...
b)\(\left(\dfrac{1}{7}x-\dfrac{2}{7}\right).\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right).\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)=0\)
=> \(\dfrac{1}{7}x-\dfrac{2}{7}=0hoặc-\dfrac{1}{5}x+\dfrac{3}{5}=hoăc\dfrac{1}{3}x+\dfrac{4}{3}=0\)
+)\(~\dfrac{1}{7}x-\dfrac{2}{7}=0\) +) \(-\dfrac{1}{5}x+\dfrac{3}{5}=0\) +) \(\dfrac{1}{3}x+\dfrac{4}{3}=0\)
\(\dfrac{1}{7}x=-\dfrac{2}{7}\) \(-\dfrac{1}{5}x=-\dfrac{3}{5}\) \(\dfrac{1}{3}x=-\dfrac{4}{3}\)
\(x=2\) \(x=3\) \(x=-4\)
Vậy...
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a 1/6x+1/10x-4/15x+1=0
(1/6+1/10-4/15)x+1=0
0x+1=0
0x=-1
x=-1/0
Vậy không có x (vì không có số nào chia cho 0)
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a)1/6x + 1/10x - 4/15x + 1=0
b)(1/7x - 2/7)(-1/5x + 3/5)(1/3x + 4/3)=0
c)(x - 1/3)(y - 1/2)(z - 5)=0 và x + 2=y + 1=z + 3
bài 49; tìm x
1, x mũ 3 + 3x mũ 2 - ( x + 3 )
2, 15x - 5 + 6x mũ 2 - 2x = 0
3, 5x - 2 - 25x mũ 2 + 10x = 0
a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)
b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)
c, \(5x-2-25x^2+10x=0\)
\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)
Tìm x
a)3x(x+5)-2(x+5)=0
b)2(x+3)-15x-5x2
c)x2-4x=5
a: =>(x+5)(3x-2)=0
=>x=-5 hoặc x=2/3
b: Đề thiếu rồi bạn
c: \(\Leftrightarrow x^2-4x-5=0\)
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
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