x²+7x+10=0
x (7x - 42 ) + 10 (7x -42 ) =0
x.( 7x - 42 ) + 10. ( 7x - 42 ) = 0
=> ( 7x - 42 ). ( x + 10 ) = 0
=> \(\orbr{\begin{cases}x+10=0\\7x-42=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-10\\x=6\end{cases}}\)
a) 7x .(2x+10)=0
b)-9x:(2x-10)=0
c) (4-x) (x+3)=0
d) (x+2023) . (x - 2024)=0
a, 7\(x\).(2\(x\) + 10) = 0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\){-5; 0}
b, - 9\(x\) : (2\(x\) - 10) = 0
- 9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
d, (\(x\) + 2023).(\(x\) - 2024) = 0
\(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-2023; 2024}
Tìm xEZ, biết
a) 7x .(2x+10)=0
b)-9x:(2x-10)=0
c) (4-x) (x+3)=0
d) (x+2023) . (x - 2024)=0
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
d, (\(x\) + 2023).(\(x\) - 2024) = 0
\(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)
Vậy \(x\in\) {-2023; 2024}
Giai phương trình sau:
a,\(x^2+3x-10=0\) b,\(3x^2-7x+1=0\)
c,\(3x^2-7x+8=0\) d,\(4x^2-12x+9=0\)
e,\(3x^2+7x+2=0\) h,\(x^2-4x+1=0\)
i,\(2x^2-6x+1=0\) j, \(3x^2+4x-4=0\)
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
giải pt:
a) (x2-3x)(x2+7x+10)=216
b) (2x2-7x+3)(2x2+x-3)+9=0
a) \(\left(x^2-3x\right)\left(x^2+7x+10\right)=216\Rightarrow x\left(x-3\right)\left(x+2\right)\left(x+5\right)=216\)
\(\Rightarrow x\left(x+2\right)\left(x-3\right)\left(x+5\right)=216\Rightarrow\left(x^2+2x\right)\left(x^2+2x-15\right)=216\)
Đặt \(t=x^2+2x\Rightarrow\) pt trở thành \(t\left(t-15\right)=216\Rightarrow t^2-15t-216=0\)
\(\Rightarrow\left(t+9\right)\left(t-24\right)=0\Rightarrow\left[{}\begin{matrix}t=-9\\t=24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+2x=-9\\x^2+2x=24\end{matrix}\right.\)
\(TH_1:x^2+2x=-9\Rightarrow x^2+2x+9=0\Rightarrow\left(x+1\right)^2+8=0\) (vô lý)
\(TH_2:x^2+2x=24\Rightarrow x^2+2x-24=0\Rightarrow\left(x-4\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) \(\left(2x^2-7x+3\right)\left(2x^2+x-3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x-1\right)\left(x-1\right)\left(2x+3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x+3\right)\left(x-1\right)\left(2x-1\right)+9=0\)
\(\Rightarrow\left(2x^2-3x-9\right)\left(2x^2-3x+1\right)+9=0\)
Đặt \(t=2x^2-3x-9\Rightarrow\) pt trở thành \(t\left(t+10\right)+9=0\)
\(\Rightarrow t^2+10t+9=0\Rightarrow\left(t+1\right)\left(t+9\right)=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-9\end{matrix}\right.\)
\(TH_1:t=-1\Rightarrow2x^2-3x-9=-1\Rightarrow2x^2-3x-8=0\)
\(\Delta=\left(-3\right)^2-4\left(-8\right).2=73\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{73}}{4}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{73}}{4}\end{matrix}\right.\)
\(TH_2:t=-9\Rightarrow2x^2-3x-9=-9\Rightarrow2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
a) x2 - 7x + 10= 0
b) 7x2 + 13x - 2 = 0
a) x2 - 7x + 10 = 0
x2 - 5x - 2x + 10 = 0
(x2 - 5x) - (2x - 10) = 0
x (x - 5) - 2 (x - 5) = 0
(x - 5) (x - 2) = 0
\(\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}}\)
\(\orbr{\begin{cases}x=0+5\\x=0+2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
Vậy x = 5 hoặc x = 2
b) tương tự
a) (x - 5)(x-2) = 0
=> x=2 , x=5
b) ( x+2)(7x-1) =0
=> x= -2 , x= 1/7
a) x.x - 7 x = - 10
x( x - 7 ) = - 10
=> x ; x -7 thuộc ước của 10
7x.(x-10)=0
7x(x-10)=0
\(\Rightarrow\orbr{\begin{cases}x-10=0\\7x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\x=0\end{cases}}}\)
Vậy x\(\in\left\{10;0\right\}\)
Đề bn ghi thiếu gì chăng?
(=)\(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\left(=\right)\orbr{\begin{cases}x=0\\x=10\end{cases}}}\left(TM\right)\)
vậy với x=0 hoặc x=10 thì ghi lại đề
#Học-tốt
Trả lời:
7x.(x - 10) = 0
=> x = 0 hoặc x - 10 = 0
=> x = 10
Vậy x = 0; x = 10
x^2+7x+10=0 tim x
Ta có:
x2 + 7x + 10 = 0
<=> x^2 + 5x + 2x + 10 = 0
<=> x(x + 5) + 2(x + 5) = 0
<=> (x+2)(x+5) = 0
<=> x+2=0 hoặc x+5=0
<=> x= -2 hoặc x= -5
Vậy x = -2; -5.
\(x^2+7x+10=0\)
\(\Leftrightarrow\left(x^2+5x\right)+\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)+2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+5=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=-2\end{cases}}}\)
giải phương trình :
a, x/4 - 3x + 11 = 5/6 - x +7x
b,x^2 - 2x = 0
c, x^2 - 7x - 10 =0
làm ơn giúp tớ với
a, x/4 - 3x + 11 = 5/6 - x +7x
\(\frac{44-11x}{4}=\frac{36x+5}{6}\Rightarrow\left(44-11x\right)6=4\left(36x+5\right)\)
\(\Rightarrow264-66x=144x+20\)
\(\Rightarrow-210x=-244\)
\(\Rightarrow x=\frac{122}{105}\)
b,x^2 - 2x = 0
=>x(x-2)=0
=>x=0 hoặc x-2=0
=>x=0 hoặc x=2
c, x^2 - 7x - 10 =0
đề có khi sai