(3x^3-8x^2+x+6):(x-2)
Dùng phương pháp đặt biến số phụ, phân tích các đa thức sau thành nhân tử
a. (x^2 + x)^2 - 2(x^2 + x) - 15
b. (x+2)(x+3)(x+4)(x+5) - 24
c. (x^2 + 8x + 7)(x^2 + 8x + 15) + 15
d. (x^2 + 3x + 1)(x^2 + 3x + 2) - 6
e. (4x+1)(12x-1)(3x+2)(x+1) - 4
f. 4(x+5)(x+6)(x+10)(x+12) - 3x^2
g. 3x^6 - 4x^5 + 2x^4 - 8x^3 + 2x^2 - 4x + 3
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
rút gọn P = 1 + x+3/x^+5x+6 : (8x^2/4x^3-8x^2 - 3/3x^2-12 - 1/x+2)
Cho hai đa thức
P(x)=3x^3+8x^2-7x-6
Q(x)=4x^3-8x^2+3x+6
a) Tính P(0); Q(-1)
b) Tính P(x)+Q(x); P(x)-Q(x)
Giải Phương trình: 13 - x / x+3 - 6x^2 + 6 / x^4 - 8x^2 - 9 - 3x + 6 / x^2 + 5x + 6 - 2 / x -3 =0
mọi người ơi giúp em với ạ ! giải chi tiết giúp em ạ
1) 3x+2/6- 3x-2/4= 15/8
2) x+2/3+x - x/3-x= 8x-6/9-x ngũ 2
mọi người ơi giúp em với ạ ! giải chi tiết giúp em ạ
1) 3x+2/6- 3x-2/4= 15/8
2) x+2/3+x - x/3-x= 8x-6/9-x ngũ 2
\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)
2, ĐKXĐ:\(x\ne\pm3\)
\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)
\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)
\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)
\(\Leftrightarrow12x+8-18x+12=45\)
\(\Leftrightarrow12x-18x=45-12-8\)
\(\Leftrightarrow-6x=25\)
\(\Leftrightarrow x=\dfrac{-25}{6}\)
Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)
\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)
\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)
\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)
\(\Leftrightarrow-x^2-x^2+3x-2x-3x-8x=-6+6\)
\(\Leftrightarrow-2x^2-10x=0\)
\(\Leftrightarrow-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;5\right\}\)
Giải pt sau:
a, 3 - 4x( 25 - 2x ) - 8x2 + x - 300
b, 2( 1 -3x )/5 - 2+ 3x/10 = 7- 3( x + 1)/4
c, 5x + 2 /6 - 8x - 1/3 = 4x + 2/5 - 5
d, 3x + 2/3 - 3x + 1/6 = 2x + 5/3
Help me
a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)
\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)
\(\Leftrightarrow-297-99x=0\)
\(\Leftrightarrow x=3\)
Vậy \(n_0\) của PT là: x=3
b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)
\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)
\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)
\(\Leftrightarrow-64-36x=250-30x\)
\(\Leftrightarrow-6x=314\)
\(\Leftrightarrow x=-\frac{157}{3}\)
Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)
c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)
\(\Leftrightarrow-3x=4x-\frac{23}{5}\)
\(\Leftrightarrow7x=\frac{23}{5}\)
\(\Leftrightarrow x=\frac{23}{35}\)
Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)
d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow x=-\frac{5}{12}\)
Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)
Rút gọn biểu thức sau:
\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{x+2}\right)\)
\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{2}\right)\)\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{3x}{3\left(4-x^2\right)}-\dfrac{1}{x+2}\right)\)\(=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{2}{x-2}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2x+4-3x-x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{-2x+6}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-2x+6}\)
\(=1+\dfrac{x-2}{-2x+6}\)
\(=\dfrac{-2x+6+x-2}{-2x+6}=\dfrac{4-x}{-2\left(x-3\right)}\)