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Bài 1 :
a, \(\left(x^2-2x+3\right)\left(x-4\right)=0\)
TH1 : \(x^2-2x+3=0\)
\(\left(-2\right)^2-4.3=4-12< 0\)vô nghiệm
TH2 : \(x-4=0\Leftrightarrow x=4\)
b, \(\left(2x^2-3x-1\right)\left(5x+2\right)=0\)
TH1 : \(\left(-3\right)^2-4.\left(-1\right).2=9+8=17>0\)
\(\Rightarrow x_1=\frac{3-\sqrt{17}}{4};x_2=\frac{3+\sqrt{17}}{4}\)
TH2 ; \(5x+2=0\Leftrightarrow x=-\frac{2}{5}\)
c, đưa về hệ đc ko ?
d, \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)=0\)
TH1 : \(x=0,74...\) ( bấm máy cx ra )
TH2 : \(\left(-1\right)^2-4.2.4< 0\)vô nghiệm
KL : vô nghiệm
Bài 2 :
a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18x-12\right)\)
\(=6x^2+21x-2x-7-6x^2+5x-6x+5-18x+12=10\)
Vậy biểu thức ko phụ thuộc vào biến
b, \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4y^4\)
\(=x^4+x^3y+x^2y^2+xy^3-yx^3-y^2x^2-y^3x-y^4-x^4y^4\)
\(=x^4-y^4-x^4y^4\)Vậy biểu thức phụ thuộc vào biến
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
Thanks bn nha !
Nhưng mik muốn câu trả lời đầy đủ hơn ạ.
MN GIÚP MIK VS !!!
b: Vì (d1)//(d3) nên a=1
hay (d1): y=x+b
Thay x=2 và y=3 vào (d1), ta được:
b+2=3
hay b=1
Ukm
It's very hard
l can't do it
Sorry!
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x^2\left(x+2\right)-x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-3\right)=0\). Làm nốt
b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-3\right)^2+\left|y+3\right|=0\)
Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
c) \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-2.\left(2x^2+x\right).2+4-1=0\)
\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)
Từ đó tính đc x
d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+5=a\), khi đó pt có dạng:
\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
a) 3x³ + 6x²y
= 3x².(x + 2y)
b) 2x³ - 6x²
= 2x².(x - 2)
c) 18x² - 20xy
= 2x.(9x - 10y)
d) xy + y² - x - y
= (xy + y²) - (x + y)
= y(x + y) - (x + y)
= (x + y)(y - 1)
e) (x²y² - 8)² - 1
= (x²y² - 8 - 1)(x²y² - 8 + 1)
= (x²y² - 9)(x²y² - 7)
= (xy - 3)(xy + 3)(x²y² - 7)
f) x² - 7x - 8
= x² - 8x + x - 8
= (x² - 8x) + (x - 8)
= x(x - 8) + (x - 8)
= (x - 8)(x + 1)
a: \(3x^3+6x^2y\)
\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)
b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)
c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)
d: \(xy+y^2-x-y\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
e: \(\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)
\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)
f: \(x^2-7x-8\)
\(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)
g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)
\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)
\(=2x\left(2x-y\right)\left(5x-3y\right)\)
h: \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1-y\right)\left(x-1+y\right)\)
i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)
\(=2x\left(x+2\right)-x^2\left(x+2\right)\)
\(=\left(x+2\right)\left(2x-x^2\right)=x\cdot\left(x+2\right)\left(2-x\right)\)
k: \(-x^2+6x-9=-\left(x^2-6x+9\right)\)
\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)
l: \(-2x^2+8xy-8y^2\)
\(=-2\left(x^2-4xy+4y^2\right)\)
\(=-2\left(x-2y\right)^2\)
m: \(3x^2+5x-3y^2-5y\)
\(=3\left(x^2-y^2\right)+5\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+3y+5\right)\)
