a 9x2-18x=0
b x(x-2)+5(2+x)=0
c 3x2-147=0
d 4x2-12x=0
e 320-5x2=0
f 2x(x-17)+(17-x)=0
Không giải phương trình, hãy tính tổng và tích các nghiệm (nếu có) của mỗi phương trình sau:
a ) 4 x 2 + 2 x − 5 = 0 b ) 9 x 2 − 12 x + 4 = 0 c ) 5 x 2 + x + 2 = 0 d ) 159 x 2 − 2 x − 1 = 0
a) Phương trình 4 x 2 + 2 x − 5 = 0
Có a = 4; b = 2; c = -5, a.c < 0
⇒ Phương trình có hai nghiệm x 1 ; x 2
Theo hệ thức Vi-et ta có:
b) Phương trình . 9 x 2 − 12 x + 4 = 0
Có a = 9; b' = -6; c = 4 ⇒ Δ 2 = ( - 6 ) 2 - 4 . 9 = 0
⇒ Phương trình có nghiệm kép x 1 = x 2 .
Theo hệ thức Vi-et ta có:
c) Phương trình 5 x 2 + x + 2 = 0
Có a = 5; b = 1; c = 2 ⇒ Δ = 1 2 − 4.2.5 = − 39 < 0
⇒ Phương trình vô nghiệm.
d) Phương trình 159 x 2 − 2 x − 1 = 0
Có a = 159; b = -2; c = -1; a.c < 0
⇒ Phương trình có hai nghiệm phân biệt x 1 ; x 2 .
Theo hệ thức Vi-et ta có:
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
2) Phân tích đa thức thành nhân tử
a) 3xy2 – 3x3 – 6xy +3x
b) 3x2 + 11x + 6
c) –x3 – 4xy2 + 4x2y +16x
d) xz – x2 – yz +2xy – y2
e) 4x2 – y2 – 6x + 3y
f) X4 – x3 – 10x2 + 2x +4
g) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
h) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik vs cần gấp!!!
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
2) Phân tích đa thức thành nhân tử
a) 3xy2 – 3x3 – 6xy +3x
b) 3x2 + 11x + 6
c) –x3 – 4xy2 + 4x2y +16x
d) xz – x2 – yz +2xy – y2
e) 4x2 – y2 – 6x + 3y
f) X4 – x3 – 10x2 + 2x +4
g) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
h) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik với mik đang cần rất gấp ạ!!!
Tìm x biết:
1,
a,3x(x+1) - 2x(x+2) = -x-1
b,2x(x-2020) - x+2020 = 0
c,(x-4)2 - 36 = 0
d,x2 + 8x - 16 = 0
e,x(x+6) - 7x - 42 = 0
f,25x2 - 16 = 0
2,
a,3x3 - 12x = 0
b,x2 + 3x - 10 = 0
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Giải các phương trình sau:
a/ (3x – 2)(4x + 5) = 0
b/ (2,3x – 6,9)(0,1x + 2) = 0
c/ (4x + 2)(x2 + 1) = 0
d/(2x + 7)(x – 5)(5x + 1) = 0
e/ (x – 1)(2x + 7)(x2 + 2) = 0
f/ (3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
Vì \(x^2+2\ge2>0\forall x\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)
Tìm x,biết
a)4x 9x2-1=0
b)(x+2)2 -(x+2)(x-3)=0
c)2x3-4x2+2x=0
d)(x-1)2-(2x+1)2=0
\(b,\Rightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Rightarrow5\left(x+2\right)=0\\ \Rightarrow x=-2\\ c,\Rightarrow2x\left(x^2-2x+1\right)=0\\ \Rightarrow2x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ d,\Rightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Rightarrow3x\left(-x-2\right)=0\\ \Rightarrow-3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
a)thiếu dấu
b)(x+2)2 -(x+2)(x-3)=0
(x+2)(x+2-x+3)=0
(x+2)5=0
x+2=0
x=-2
c)2x3-4x2+2x=0
2x(x2-2x+1)=0
2x(x-1)2
suy ra 2 trường hợp
x=0
x-1=0=>x=1
d)(x-1)2-(2x+1)2=0
(x-1-2x-1)(x-1+2x+1)=0
(x-2)3x=0
x=0
x=2
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
Giúp mik vs cần gấp!!!
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương
Bài 5: Giải các phương trình sau:
a. (3x - 1)2 - (x + 3)2 = 0
b. x3 = \(\dfrac{x}{49}\)
c. x2 - 7x + 12 = 0
d. 4x2 - 3x -1 = 0
e. x3 - 2x - 4 = 0
f. x3 + 8x2 + 17x +10 = 0
g. x3 + 3x2 + 6x + 4 = 0
h. x3 - 11x2 + 30x = 0
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
*Cách khác:
a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)
Tìm X biết:
a/ 5x (x - 2000) – x + 2000 = 0
b/ x3 – 13x = 0
c/ 3x2 - 6x =0
d/ x(x- 5) +3(x-5)=0
e/9x2 – 4 =0
giúp mình vs thx mn
\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)