x^2 - x - 20 = 0
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\(x^2-x-20=0\)
\(\Leftrightarrow x^2-5x+4x-20=0\)
\(\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}}\)
cn lại lm tg tự nha bn
=.= hok tốt!!
`Answer:`
\(x^2-x-20=0\)
\(\Leftrightarrow x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{81}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{81}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\left(\frac{9}{2}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{9}{2}\\x-\frac{1}{2}=-\frac{9}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
\(x^2+80x-20=0\)
\(\Leftrightarrow x^2+2.40x+1600-1620=0\)
\(\Leftrightarrow\left(x+40\right)^2-\sqrt{1620}=0\)
\(\Leftrightarrow\left(x+40\right)^2=18\sqrt{5}\)
\(\Leftrightarrow\orbr{\begin{cases}x+40=18\sqrt{5}\\x+40=-18\sqrt{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=18\sqrt{5}-40\\x=-18\sqrt{5}-40\end{cases}}\)
\(x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x.\left(x-1\right)+6.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+6=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-6\\x=1\end{cases}}\)
a: Ta có: \(x^2-8x+20\)
\(=x^2-8x+16+4\)
\(=\left(x-4\right)^2+4>0\forall x\)
b: Ta có: \(-x^2+6x-19\)
\(=-\left(x^2-6x+19\right)\)
\(=-\left(x^2-6x+9+10\right)\)
\(=-\left(x-3\right)^2-10< 0\forall x\)
a) Dễ thấy x = 0 thuộc tập xác định của hàm số.
\(f\left( 0 \right) = {0^2} + 1 = 1\)
Ta có: \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{x^2} + 1} \right) = {0^2} + 1 = 1\)
\(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {1 - x} \right) = 1 - 0 = 1\)
Vì \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 1\) nên \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1 = f\left( 0 \right)\).
Vậy hàm số liên tục tại điểm \(x = 0\).
b)Dễ thấy x = 1 thuộc tập xác định của hàm số.
\(f\left( 1 \right) = {1^2} + 2 = 3\)
Ta có: \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} + 2} \right) = {1^2} + 2 = 3\)
\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} x = 1\)
Vì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).
Vậy hàm số không liên tục tại điểm \(x = 1\).
a) x^2 +3x-2x-6=0
x^2 + x = 6
x^2 + 0.5x + 0.5x = 6
x (x + 0.5) + 0.5 (x + 0.5) =5.75
(x+0.5)^2 = 5.75
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
`1/2 : x-5/6 =-2/3`
`=> 1/2 : x=-2/3 +5/6`
`=> 1/2 : x= -4/6 +5/6`
`=> 1/2 : x=1/6`
`=>x=1/2:1/6`
`=>x= 1/2 xx 6`
`=>x= 6/2`
`=>x=3`
Vậy `x=3`
__
`20% . x +5/8 -x . 0,5 =11/20`
`=> 20/100 . x + 5/8 - x . 5/10=11/20`
`=> 1/5 . x+5/8 - x. 1/2 =11/20`
`=> (1/5 -1/2) . x+5/8=11/20`
`=>-3/10 . x+ 5/8 =11/20`
`=> -3/10 . x=11/20 -5/8`
`=>-3/10 .x=-3/40`
`=> x= -3/40 : (-3/10)`
`=> x=-3/40 xx (-10/3)`
`=>x= 1/4`
Vậy `x=1/4`
` @ ` \(\text{Nguyễn Hoàng Duy Khánh}\)
3(2x+y)-2(3x-2y)=3.19-11.2
6x+3y-6x+4y=57-22
7y=35
y=5
thay vào :
2x+y=19
2x+5=19
2x=14
x=7
2/ x2+21x-1x-21=0
x(x+21)-1(x+21)=0
(x+21)(x-1)=0
TH1 x+21=0
x=-21
TH2 x-1=0
x=1
vậy x = {-21} ; {1}
3/ x4-16x2-4x2+64=0
x2(x2-16)-4(x2-16)=0
(x2-16)-(x2-4)=0
TH1 x2-16=0
x2=16
<=>x=4;-4
TH2 x2-4=0
x2=4
x=2;-2
Bài 1 :
\(\hept{\begin{cases}2x+y=19\\3x-2y=11\end{cases}\Leftrightarrow\hept{\begin{cases}4x+2y=38\\3x-2y=11\end{cases}\Leftrightarrow\hept{\begin{cases}7x=49\\2x+y=19\end{cases}}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7\\2x+y=19\end{cases}}\)Thay vào x = 7 vào pt 2 ta được :
\(14+y=19\Leftrightarrow y=5\)Vậy hệ pt có một nghiệm ( x ; y ) = ( 7 ; 5 )
Bài 2 :
\(x^2+20x-21=0\)
\(\Delta=400-4\left(-21\right)=400+84=484\)
\(x_1=\frac{-20-22}{2}=-24;x_2=\frac{-20+22}{2}=1\)
Bài 3 : Đặt \(x^2=t\left(t\ge0\right)\)
\(t^2-20t+64=0\)
\(\Delta=400+4.64=656\)
\(t_1=\frac{20+4\sqrt{41}}{2}\left(tm\right);t_2=\frac{20-4\sqrt{41}}{2}\left(ktm\right)\)
Theo cách đặt : \(x^2=\frac{20+4\sqrt{41}}{2}\Rightarrow x=\sqrt{\frac{20+4\sqrt{41}}{2}}=\frac{\sqrt{20\sqrt{2}+4\sqrt{82}}}{2}\)
a, 20 . x - 3 = 0
=> x - 3 = 0 : 20
=> x - 3 = 0
=> x = 0 + 3
=> x = 3
b, ( x - 1 ) . ( x - 2 ) = 0
TH 1 :
x-1 = 0
=> x = 0 + 1
=> x = 1
TH2 :
x - 2 = 0
=> x = 0 + 2
=> x = 2
Vậy x = 1 hoặc x = 2
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