giải các pt sau
a\ \(2x^2-x=3-6x\)
b\ \(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right)x^2\)
Giải các pt sau
a, \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)\)=0
b,\(\left(2x-1\right)\left(x-5\right)\left(x^2+3\right)\)=0
c,\(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
d,\(\left(2x+3\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
a); b) Do tích = 0
=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)
=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)
a; *x-1=0 <=>x=1
*2x+5=0 <=>x=-2,5
*x2+2=0 <=> ko có x
b; tương tự a
a/ \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)
Vì \(x^2\ge0\Rightarrow x^2+2\ge2>0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
Bằng cách phân tích vế trái thành nhân tử, giải các PT sau:
a) \(2x.\left(x-3\right)+5\left(x-3\right)\)
b) \(\left(x^2-4\right)+\left(x-2\right).\left(3-2x\right)=0\)
c) \(x^3-3x^2+3x-1=0\)
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
giải các phương trình sau
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\)16
\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)=3\)
giải pt sau
a)\(\left(x-2\right)\left(x-3\right)+2x=\left(x-2\right)^2-2\)
b) \(\left(x-1\right)^2+3x\left(x-1\right)+7=\left(2x-1\right)^2+5\left(x-3\right)\)
c)\(5\left(x^1-2x-1\right)+2\left(3x-2\right)=5\left(x+1\right)^2\)
d)\(\left(x-1\right)\left(x^2+x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\)
Lời giải:
a)
\((x-2)(x-3)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)(x-2-1)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)^2-(x-2)+2x=(x-2)^2-2\)
\(\Leftrightarrow x+4=0\Rightarrow x=-4\)
b)
\((x-1)^2+3x(x-1)+7=(2x-1)^2+5(x-3)\)
\(\Leftrightarrow (x-1)^2+3x(x-1)+7=x^2+(x-1)^2+2x(x-1)+5(x-3)\)
\(\Leftrightarrow x(x-1)+7=x^2+5(x-3)\)
\(\Leftrightarrow 6x=22\Rightarrow x=\frac{11}{3}\)
c)
\(5(x^2-2x-1)+2(3x-2)=5(x+1)^2=5(x^2-2x+1)\)
\(\Leftrightarrow -5+2(3x-2)=5\)
\(\Leftrightarrow 3x-2=5\Rightarrow x=\frac{7}{3}\)
d)
\((x-1)(x^2+x+1)-2x=x(x-1)(x+1)=x(x^2-1)\)
\(\Leftrightarrow x^3-1-2x=x^3-x\Leftrightarrow -1-x=0\Rightarrow x=-1\)
Bài `1`: Rút gọn các biểu thức sau:
\(a)4x^2\left(5x^2+3\right)-6x\left(3x^3-2x+1\right)-5x^3\left(2x-1\right)\)
\(b)\dfrac{3}{2}x\left(x^2-\dfrac{2}{3}x+2\right)-\dfrac{5}{3}x^2\left(x+\dfrac{6}{5}\right)\)
Bài `2`: Thực hiện các phép nhân sau:
\(a)\left(x^2-x\right)\cdot\left(2x^2-x-10\right)\)
\(b)\left(0,2x^2-3x\right)\cdot5\left(x^2-7x+3\right)\)
\(c)6x^2\cdot\left(2x^3-3x^2+5x-4\right)\)
\(d)\left(-1,2x^2\right)\cdot\left(2,5x^4-2x^3+x^2-1,5\right)\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Tính:
\(a)\left(-2x^2\right)\cdot\left(3x-4x^3+7-x^2\right)\)
\(b)\left(x+3\right)\cdot\left(2x^2-3x-5\right)\)
\(c)\left(-6x^5+7x^4-6x^3\right):3x^3\)
\(d)\left(9x^2-4\right):\left(3x+2\right)\)
\(e)\left(2x^4-13x^3+15x^2+11x-3\right):\left(x^2-4x-3\right)\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
Bài 1: giải hệ pt
\(\left\{{}\begin{matrix}x+2y=1\\2x^{2^{ }}-5xy=48\end{matrix}\right.\)
bài 2: giải các pt sau:
a/ \(\left(x^2-1\right)^2-4\left(x^2-1\right)=5\)
b/\(\left(x+2\right)^2-3x-5=\left(1-x\right)\left(1+x\right)\)
c/ \(\left(x^2-3x+4\right)\left(x^2-3x+2\right)=3\)
Bài 1:
\(\left\{{}\begin{matrix}x+2y=1\\2x^2-5xy=48\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1-2y\left(1\right)\\2x^2-5xy=48\left(2\right)\end{matrix}\right.\)
Thay (1) vào (2)\(\Leftrightarrow2\left(1-2y\right)^2-5\left(1-2y\right)y=48\Leftrightarrow2\left(1-4y+4y^2\right)-5y+10y^2=48\Leftrightarrow2-8y+8y^2-5y+10y^2=48\Leftrightarrow18y^2-13y-46=0\Leftrightarrow\left(y-2\right)\left(18y+23\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y=2\\y=-\frac{23}{18}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\x=\frac{32}{9}\end{matrix}\right.\)
Vậy (x;y)={(\(-3;2\));(\(\frac{32}{9};-\frac{23}{18}\))}
Bài 2:
a) Đặt a=x2-1(a\(\ge-1\))
Vậy pt\(\Leftrightarrow a^2-4a=5\Leftrightarrow a^2-4a-5=0\Leftrightarrow\left(a-5\right)\left(a+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=5\\a=-1\end{matrix}\right.\)(tm)
TH1: a=5\(\Leftrightarrow x^2-1=5\Leftrightarrow x^2=6\Leftrightarrow x=\pm\sqrt{6}\)
TH2: a=-1\(\Leftrightarrow x^2-1=-1\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Vậy S={\(-\sqrt{6};0;\sqrt{6}\)}
b) \(\left(x+2\right)^2-3x-5=\left(1-x\right)\left(1+x\right)\Leftrightarrow x^2+4x+4-3x-5=1-x^2\Leftrightarrow2x^2+x-2=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{-1+\sqrt{17}}{4}\\x=\frac{-1-\sqrt{17}}{4}\end{matrix}\right.\)
Vậy S={\(\frac{-1+\sqrt{17}}{4};\frac{-1-\sqrt{17}}{4}\)}
c) Đặt a=\(x^2-3x+2\)
Vậy pt\(\Leftrightarrow\left(a+2\right)a=3\Leftrightarrow a^2+2a-3=0\Leftrightarrow\left(a-1\right)\left(a+3\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)(tm)
TH1:\(a=1\Leftrightarrow x^2-3x+2=1\Leftrightarrow x^2-3x+1=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{matrix}\right.\)
TH2: a=-3\(\Leftrightarrow x^2-3x+2=-3\Leftrightarrow x^2-3x+5=0\)(vô nghiệm)
Vậy S=\(\left\{\frac{3+\sqrt{5}}{2};\frac{3-\sqrt{5}}{2}\right\}\)