Bài 1
a. x. (x+2)=0
b. x÷1/2=3,5÷4/6
c. 1,5÷(2x)=1/2-3/4
Những câu hỏi liên quan
Tìm x:
a)5x(x-2)-2x+4=0
b)2x(x+1)-(x-2)^2=6
c)2x^2+7x-9=0
\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)
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18 tháng 8 2021 lúc 10:04
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
18 tháng 8 2021 lúc 11:17
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
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Tìm x
a,(7x+4)^2-(7x+4)(7x-4)=0
b, 5( x + 3 )( x - 3 ) + ( 2x + 3 )^2+(x-6)^=10
c, (x + 1)^3 + (x – 2)^3 – 2x^2 (x – 1,5) = 3
d,( x + 2)(x^2 – 2x + 4)(x – 2)(x^2 + 2x + 4) = – 65
e, 4x^2 + 4x – 5 = 2
f,16x^2 – 9(x + 1)^2 = 0
Các bạn giúp mình vs mai mình phải nộp rùii
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
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Bài 2: giải phương trình sau a) X^4-x^2-20b) (x+1)^4-x^2+2)^20c)3x^2-2x-80Bài 3: giải phương trình sau a) x^3-0,250b) x^4+2x^3+x^20c) x^3-10d) 6x^2-7x+20Mong có người giải giùm xin kẻm ơn :
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Bài 2: giải phương trình sau
a) \(X^4\)-\(x^2\)-2=0
b) (x+1)\(^4\)-x\(^2\)+2)\(^2\)=0
c)3x\(^2\)-2x-8=0
Bài 3: giải phương trình sau
a) x\(^3\)-0,25=0
b) x\(^4\)+2x\(^3\)+x\(^2\)=0
c) x\(^3\)-1=0
d) 6x\(^2\)-7x+2=0
Mong có người giải giùm xin kẻm ơn :>
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
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13 tháng 9 2021 lúc 17:12
bài 3: tìm x , biết:
a)(x+3).(2x-1)-(x-3).(x+1)=0
b)(x+4).(2x-3)-3.(x-2).(x+2)=0
c)x.(x-5).(x+5)-(x+2).(x2-2x+4)=17
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
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Bài 1:
a) (x-1/3)^2=0
b) (x-4)^2=16
c) (2x-1)^3= -8
Bài 2:
a) (-1/30)^0
b) (3 1/4)^2
c) (-1 3/4)^2
d) (3/7)^20 : (9/49)^6
e) 3^2.5^2 .(2/3)^2
\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)
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Bài 2: Tìm x, biết:
a/ 1,5 - |x - 0,3| = 0
b/| 2𝑥+3|−24=−9
c/ |𝑥+13|—12=0
d/ (-4,6)+ x =(-3,5)
e/ |1,5𝑥|−|−0,6| = -2+|0,4|
d: Ta có: (-4,6)+x=-3,5
nên x=-3,5+4,6
hay x=1,1
a: Ta có: \(1.5-\left|x-0.3\right|=0\)
\(\Leftrightarrow\left|x-0.3\right|=1.5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0.3=1.5\\x-0.3=-1.5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.2\end{matrix}\right.\)
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bài 1 tìm x
a)6x^2-72x=0
b)-2x^4+16x=0
c)x(x-5)-(x-3)^2=0
d)(x-2)^3-(x-2)(x^2+2x+4)=0
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
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