Ôn tập: Phân thức đại số

các bn giúp mik với!! vài câu cx được

a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)

=>-12x-4=2x-10

=>-14x=-6

hay x=3/7

b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)

=>15x-9-10x+2=-4

=>5x-7=-4

=>5x=3

hay x=3/5(loại)

c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)

\(\Leftrightarrow x^2+3x-1=x^2-x+1\)

=>4x=2

hay x=1/2(nhận)

\(A=\dfrac{x^5-5x^2+4x}{30}=\dfrac{x\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)}{30}\)

Vì x;x-1;x-2;x+1;x+2 là 5 số liên tiếp

nên \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)⋮5!=120\)

=>A luôn là số nguyên

1) ở đây : https://hoc24.vn/hoi-dap/question/637404.html

2) pt \(\Leftrightarrow\left(9x^2+18x-16\right)\left(x^2+2x+1\right)=-16\)

\(\Leftrightarrow9x^4+36x^3+29x-14x=0\)

\(\Leftrightarrow x\left(9x^3+18x^2+18x^2+36x-7x-14\right)=0\)

\(\Leftrightarrow x\left(9x^2\left(x+2\right)+18x\left(x+2\right)-7\left(x+2\right)\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(9x^2+18x-7\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(9x^2-3x+21x-7\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(3x\left(3x-1\right)+7\left(3x-1\right)\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(3x+7\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=\dfrac{-7}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\) vậy ............................................................................................................

a) \(36x^2-49=0\)

\(\Leftrightarrow\left(6x\right)^2-7^2=0\)

\(\Leftrightarrow\left(6x-7\right)\left(6x+7\right)=0\)

\(TH_1:6x-7=0\) \(TH_2:6x+7=0\)

\(\Leftrightarrow6x=7\) \(\Leftrightarrow6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\) \(\Leftrightarrow x=-\dfrac{7}{6}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{7}{6};-\dfrac{7}{6}\right\}\)

Bài 2

a) 36x²-49=0

⇔ (6x)²-49=0

⇔(6x-7).(6x+7)=0

TH1: 6x-7=0 TH2: 6x+7=0

⇔6x=7 ⇔6x=-7

⇔x=7/6 ⇔x=-7/6

\(f\left(100\right)\Leftrightarrow x=100\)

\(\Rightarrow x+1=101\left(1\right)\)

Thay (1) vào ta được

\(f\left(100\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^2-\left(x+1\right)x+25\)

\(f\left(100\right)=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...+x^2-x^2-x+25\)

\(f\left(100\right)=-x+25\)

\(f\left(100\right)=-100+25\)

\(f\left(100\right)=-75\)

a: ĐKXĐ: \(x\notin\pm\sqrt{10}\)

b: \(M=\left(\dfrac{5x+2}{x^2-10}+\dfrac{5x-2}{x^2+10}\right)\cdot\dfrac{x^4-100}{x^2+4}\)

\(=\dfrac{5x^3+50x+2x^2+20+5x^3-50x-2x^2+20}{x^4-100}\cdot\dfrac{x^4-100}{x^2+4}\)

\(=\dfrac{10x^3+40}{x^2+4}\)

\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{2x+1}{-\left(2x+1\right)}-\dfrac{2x-1}{-\left(2x-1\right)}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{2x}{5x-5}\)

ta có : \(x+3+\dfrac{4-3a^2}{a^2-9}=\dfrac{5}{2a^2+6a}\)

\(\Leftrightarrow x+3=\dfrac{5}{2a^2+6a}-\dfrac{4-3a^2}{a^2-9}\)

\(\Leftrightarrow x=\dfrac{6a^3-3a-15}{2a\left(a+3\right)\left(a-3\right)}-3=\dfrac{6a^3-3a-15-3.2a\left(a^2-9\right)}{2a\left(a+3\right)\left(a-3\right)}\)

\(\Leftrightarrow x=\dfrac{6a^3-3a-15-6a^3+54a}{2a\left(a+3\right)\left(a-3\right)}=\dfrac{51a-15}{2a\left(a^2-9\right)}\)

ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\)

\(=23+2\left(xy+yz+zx\right)=49\Rightarrow xy+yz+zx=13\)

rồi bn có gắn qui đồng nó thế vào là o ke :( mk qui vài mà nó dài quá thôi bỏ luôn