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Giải phương trình:
A)\(3X^2+12X-66\)bằng \(0\)
B)\(9X^2-30X+225\)bằng \(0\)
C)\(2X^2-6X+1\)bằng \(0\)
D)\(3X^2-7X+8\)bằng \(0\)
E)\(3X^2-7X+1\)bằng \(0\)
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
22 tháng 2 2020
a)
\(3x^2+12x-66=0\)
\(\Leftrightarrow x^2+4x-22=0\)
\(\Leftrightarrow x^2+4x+4=26\Leftrightarrow (x+2)^2=26\)
\(\Rightarrow x+2=\pm \sqrt{26}\Rightarrow x=-2\pm \sqrt{26}\)
b)
\(9x^2-30x+225=0\)
\(\Leftrightarrow (3x)^2-2.3x.5+25+200=0\)
\(\Leftrightarrow (3x-5)^2=-200< 0\) (vô lý nên pt vô nghiệm)
c)
\(x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x(x-2)+5(x-2)=0\Leftrightarrow (x+5)(x-2)=0\)
\(\Rightarrow x=-5\) hoặc $x=2$
AH
Akai Haruma
Giáo viên
22 tháng 2 2020
d)
$3x^2-7x+1=0$
$\Leftrightarrow 3(x^2-\frac{7}{3}x)+1=0$
$\Leftrightarrow 3(x^2-\frac{7}{3}x+\frac{7^2}{6^2})=\frac{37}{12}$
$\Leftrightarrow 3(x-\frac{7}{6})^2=\frac{37}{12}$
$\Leftrightarrow (x-\frac{7}{6})^2=\frac{37}{36}$
$\Rightarrow x-\frac{7}{6}=\frac{\pm \sqrt{37}}{6}$
$\Rightarrow x=\frac{7\pm \sqrt{37}}{6}$
e)
$3x^2+7x+2=0$
$\Leftrightarrow 3(x^2+\frac{7}{3}x+\frac{7^2}{6^2})=\frac{25}{12}$
$\Leftrightarrow 3(x+\frac{7}{6})^2=\frac{25}{12}$
$\Leftrightarrow (x+\frac{7}{6})^2=\frac{25}{36}$
$\Rightarrow x+\frac{7}{6}=\pm \frac{5}{6}$
$\Rightarrow x=\frac{-1}{3}$ hoặc $x=-2$
NA
13 tháng 6 2017
a)
a)
=> 3(x + 2)2 - 12 - 66 = 0
=> 3(x + 2)2 - 78 = 0
=> 3(x + 2)2 = 78
=> (x + 2)2 = 26
=> x = \(\sqrt{26}-2\)
b)
=> (3x - 5)2 - 25 + 225 = 0
=> (3x - 5)2 + 200 = 0
=> (3x - 5)2 = -200
9x2 - 30x + 225 không có ngiệmc)=> (x + 1,5)2 - 2,25 - 10 = 0
=> (x + 1,5)2 - 12,25 = 0
=> (x + 1,5)2 = 12, 25
=> x + 1,5 = 3,5
=> x = 2
d)=> 3(x - \(\dfrac{7}{6}\))2 - \(\dfrac{49}{12}\) + 1 = 0
=> 3(x - \(\dfrac{7}{6}\))2 - \(\dfrac{37}{12}\) = 0
=> 3(x - \(\dfrac{7}{6}\))2 = \(\dfrac{37}{12}\)
=> (x - \(\dfrac{7}{6}\))2 = \(\dfrac{37}{36}\)
=> x = \(\dfrac{\sqrt{37}}{6}+\dfrac{7}{6}=\dfrac{\sqrt{37}+7}{6}\)
e)
=> 3(x - \(\dfrac{7}{6}\))2 - \(\dfrac{49}{12}\)+ 8 = 0
=> 3(x - \(\dfrac{7}{6}\))2 + \(\dfrac{47}{12}\) = 0
=> 3(x - \(\dfrac{7}{6}\))2 = \(-\dfrac{47}{12}\)
KL : Không có ngiệm
KN
10 tháng 2 2020
a) \(3x^2+12x-66=0\)
Ta có \(\Delta=12^2+4.3.66=936,\sqrt{\Delta}=6\sqrt{26}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-12+6\sqrt{26}}{6}=-2+\sqrt{26}\\x=\frac{-12-6\sqrt{26}}{6}=-2-\sqrt{26}\end{cases}}\)
b) \(9x^2-30x+225=0\)
Ta có \(\Delta=33^2-4.9.225=-7011\)
\(\Delta< 0\)nên pt vô nghiệm
c) \(x^2+3x-10=0\)
Ta có \(\Delta=3^2+4.10=49,\sqrt{\Delta}=7\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-3+7}{2}=2\\x=\frac{-3-7}{2}=-5\end{cases}}\)
d) \(3x^2-7x+1=0\)
Ta có \(\Delta=7^2-4.3.1=37,\sqrt{\Delta}=\sqrt{37}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7+\sqrt{37}}{6}\\x=\frac{7-\sqrt{37}}{6}\end{cases}}\)
25 tháng 1 2021
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: \(x\in\varnothing\)
TN
12 tháng 3 2020
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
3 tháng 4 2020
a) ( 3.x + 1 ) . ( 7.x + 3 ) = (5.x-7 ) . ( 3.x + 1 )
<=> ( 3.x + 1 ) . ( 7.x + 3 ) - ( 5.x - 7) . ( 3.x + 1 ) = 0
<=> ( 3.x + 1 ) . ( 7.x + 3 - 5.x + 7 ) = 0
<=> ( 3.x + 1 ) . ( 2.x + 10 ) = 0
<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)
Vậy x = { \(\frac{-1}{3};-5\)}
b) x2 + 10.x + 25 - 4.x . ( x + 5 ) = 0
<=> ( x + 5 )2 -4.x . (x + 5 ) = 0
<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0
<=> ( x + 5 ) . ( 5 - 3.x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\5-3.x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)
Vậy x = \(\left\{\frac{5}{3};-5\right\}\)
c) (4.x - 5 )2 - 2. ( 16.x2 -25 ) = 0
<=> ( 4.x-5)2 -2 .( 4.x-5) .( 4.x + 5 ) = 0
<=> ( 4.x -5 )2 - ( 8.x+ 10 ) . ( 4.x -5 ) = 0
<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0
<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0
<=> \(\orbr{\begin{cases}4.x-5=0\\-4.x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}}\)
Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)
d) ( 4.x + 3 )2 = 4. ( x2 - 2.x + 1 )
<=> 16.x2 + 24.x + 9 - 4.x2 + 8.x - 4 = 0
<=> 12.x2 + 32.x + 5 =0
<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0
<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)
Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
e) x2 -11.x + 28 = 0
<=> x2 -4.x - 7.x + 28 = 0
<=> ( x - 7 ) . ( x - 4 ) = 0
<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
Vậy x = { 4 ; 7 }
f ) 3.x.3 - 3.x2 - 6.x = 0
<=> 3.x. ( x2 -x - 2 ) = 0
<=> 3.x. ( x - 2 ) . ( x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\([x=0\) \([x=0\)
( Lưu ý :Lưu ý này không cần ghi vào vở : Chị nối 2 ý đó làm 1 nha cj ! )
Vậy x = { 2 ; -1 ; 0 }