Thực hiện phép tính
\(a,\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(b,\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(c,\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(d,\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
\(e,\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(f,\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(g,\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(h,\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
Bài 2 . Thực hiện phép tính
a)\(6x^3\)\(\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)\)\(-2x^5\)\(-x^3\)
b)\(\left(x-3\right)\left(x^2+3x-2\right)\)
c)\(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
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: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
1.rút gọn biểu thuc P=\(\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}\) với x\(\ne-3vàx\ne3\)
2.thực hiện phép tính \(\left(2x^4-3x^3-3x^2+6x-1\right):\left(x^2-2\right)\)
\(\left(15x^4y^6-12^3y^4-18x^2y^3\right):\left(-6x^2y^2\right)\)
thực hiện phép tính
\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(\dfrac{15x-11}{x^2+2x-3}-\dfrac{3x-2}{x-1}-\dfrac{2x+3}{3+x}\)
\(\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(\Leftrightarrow\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(ĐKXĐ:x\ne1\)
\(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{(1+2x)\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\dfrac{6\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)
\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)
\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)
\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)
\(\Rightarrow-4x^2-8x\)
⇒-4x(x-4)
Giải phương trình
\(a,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(b,\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(c,3\left(x-1\right)+3=5x\)
\(d,\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(e,\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\)
\(f,\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
Em mới học về pt nên chưa quen lắm mọi người giúp e với ạ !Nguyễn Việt Lâm Quản lý
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
e) Ta có: \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\)
\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\)
mà \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
f) Ta có: \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
mà \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
Bài 1 : Thực hiện phép tính:
a, \(\dfrac{6}{7}.[\left(\dfrac{-7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2]\)
Bài 2 : Tìm x :
a , \(\dfrac{2}{3}:x=1,4-\dfrac{12}{5}\)
b , \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
Bài 3 : Lớp 6A có 40 h/s . Cuối năm xếp loại giỏi , khá , TB . Biết số h/s giỏi bằng \(\dfrac{1}{4}\)
số h/s cả lớp . Số h/s khá bằng \(\dfrac{3}{5}\)số h/s còn lại
Tính số h/s giỏi , khá , TB?
Bài 4 : Lớp 6B có h/s xếp loại giỏi , khá , TB , yếu . Số h/s giỏi chiếm \(\dfrac{1}{6}\) , số h/s khá chiếm \(\dfrac{1}{3}\)
số h/s cả lớp . Số h/s TB chiếm \(\dfrac{7}{8}\) số h/s giỏi . Còn lại là 17 h/s yếu . Tính số h/s lớp 6B
2:
a: =>2/3:x=1,4-2,4=-1
=>x=-2/3
b: =>x/5=25/30-19/30=6/30=1/5
=>x=1
3:
Số học sinh giỏi là 40*1/4=10 bạn
Số học sinh khá là 30*3/5=18 bạn
Số học sinh TB là 30-18=12 bạn
Thực hiện phép tính sau
\(\dfrac{3-3x}{\left(1+x\right)^2}:\dfrac{6x^2-6}{x+1}\)
\(=\dfrac{-3\left(x-1\right)}{\left(1+x\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}\left(x\ne\pm1\right)\\ =\dfrac{-1}{2\left(x+1\right)^2}\)
Bài 1 giải phương trình:
a) (4x2+4x+1)-x2=0
b) x2-2x+1=4
c) x2-5x+6=0
Bài 2: giải phương trình
a) \(\dfrac{2x-5}{x+5}\)= 3
b) \(\dfrac{5}{3x+2}\)= 2x-1
c) \(\dfrac{x^2-6}{x}\)= x+\(\dfrac{3}{2}\)
d) \(\dfrac{1}{x-2}\)+3= \(\dfrac{x-3}{2-x}\)
e) \(\dfrac{3x-2}{x+7}\)=\(\dfrac{6x+1}{2x-3}\)
f) \(\dfrac{x-2}{x+2}\) - \(\dfrac{3}{x-2}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
2f.
ĐKXĐ: $x\neq \pm 2$
PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$
$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$
$\Leftrightarrow x^2-4x+4-3x-6=2x-22$
$\Leftrightarrow x^2-7x-2=2x-22$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Leftrightarrow x-4=0$ hoặc $x-5=0$
$\Leftrightarrow x=4$ hoặc $x=5$ (tm)
