\(2+\frac{x}{5}-0,5x=1-\frac{2x}{4}+0,25\)

\(2+\frac{x}{5}-\frac{1}{2}x=1+0,25-\frac{x}{2}\)

\(2+\frac{x}{5}-\frac{x}{2}=1,25-\frac{x}{2}\)

\(\frac{x}{5}-\frac{x}{2}+\frac{x}{2}=-2+1,25\)

\(\frac{2x-5x+5x}{10}=0,75\)

\(\frac{2x}{5}=0,75\)

\(2x=0,75.5\)

\(2x=3,75\)

\(x=\frac{3,75}{2}=\frac{15}{8}\)

Vậy \(x=\frac{15}{8}\)

e) \(x^2-6x+9=45-5x^2\)

\(\Leftrightarrow6x^2-6x-36=0\)

\(\Leftrightarrow x^2-x-6=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

f) \(\left(3x-2\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow\left(3x-2\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(2x-7\right)\left(4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{4}{3}\end{cases}}\)

g) \(\left(x^2-6x\right)^2+14\left(x-3\right)^2=81\)

\(\Leftrightarrow x^4-12x^3+36x^2+14\left(x^2-6x+9\right)=81\)

\(\Leftrightarrow x^4-12x^3+50x^2-84x+45=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(11x^3-11x^2\right)+\left(39x^2-39x\right)-\left(45x-45\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-11x^2+39-45\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-3x^2\right)-\left(8x^2-24x\right)+\left(15x-45\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-8x+15\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)^2\left(x-5\right)=0\)

\(\Rightarrow x\in\left\{1;3;5\right\}\)

h) \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x+5\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-5x+5=1\\x^2-5x+5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)\left(x-4\right)=0\\\left(x-2\right)\left(x-3\right)=0\end{cases}}\)

\(\Rightarrow x\in\left\{1;2;3;4\right\}\)

e, \(x^2-6x+9=45-5x^2\)

\(\Leftrightarrow\left(x-3\right)^2=5\left(9-x^2\right)\)

\(\Leftrightarrow\left(x-3\right)^2=5\left(3-x\right)\left(x+3\right)\)

\(\Leftrightarrow\left(x-3\right)+5\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[5+\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+8\right)=0\Leftrightarrow x=3;-8\)

f, \(\left(3x-2\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow\left(3x-2\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(3x-2-x-5\right)\left(3x-2+x+5\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(4x+3\right)=0\Leftrightarrow x=\frac{7}{2};-\frac{3}{4}\)

\(x+y+z=7\Rightarrow z=7-x-y\Rightarrow xy+z-6=xy+7-x-y-6=xy-x-y+1\)

\(=\left(x-1\right)\left(y-1\right)\)

Tương tự: \(yz+x-6=\left(y-1\right)\left(z-1\right);zx+y-6=\left(z-1\right)\left(x-1\right)\)

Viết lại: \(H=\frac{1}{\left(x-1\right)\left(y-1\right)}+\frac{1}{\left(y-1\right)\left(z-1\right)}+\frac{1}{\left(z-1\right)\left(x-1\right)}\)

\(=\frac{x-1+y-1+z-1}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=\frac{x+y+z-3}{xyz-\left(xy+yz+zx\right)+x+y+z-1}\)

\(=\frac{7-3}{3-13+7-1}=-1\)(Từ gt tính được \(xy+yz+zx=13\))

Ta có :

\(xy+yz+zx\)= \(\frac{\left(x+y+z\right)^2-x^2-y^2-z^2}{2}\)= \(\frac{7^2-23}{2}\)= \(13\)

Ta lại có :

\(xy+z-6=xy+z+1-x-y-z\)= \(\left(x-1\right)\left(y-1\right)\)

\(\Rightarrow A=\)\(\frac{1}{\left(x-1\right)\left(y-1\right)}\)\(+\)\(\frac{1}{\left(y-1\right)\left(z-1\right)}\)\(+\)\(\frac{1}{\left(z-1\right)\left(x-1\right)}\)

\(=\)\(\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}\)

\(=-1\)

a) Ta có: \(x^2+2y^2+2z^2-2xy-2yz-2z=4\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2z+1\right)=5\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-1\right)^2=5\)

Mà \(5=0^2+1^2+2^2\) nên ta có dễ dàng xét được các TH

Làm tiếp nhé!

b) Ta có: \(x^2+13y^2-6xy=100\)

\(\Leftrightarrow\left(x^2-6xy+9y^2\right)+4y^2=100\)

\(\Leftrightarrow\left(x-3y\right)^2=100-4y^2\)

Mà \(\hept{\begin{cases}\left(x-3y\right)^2\ge0\\100-4y^2\le100\end{cases}}\Rightarrow0\le100-4y^2\le100\)

\(\Rightarrow y\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5\right\}\)

Ta có các TH sau:

Nếu \(y=0\Rightarrow x^2=100\Rightarrow x=\pm10\)

Nếu \(y=\pm3\Leftrightarrow\orbr{\begin{cases}\left(x-9\right)^2=64\\\left(x+9\right)^2=64\end{cases}}\Rightarrow x\in\left\{17;1;-17;-1\right\}\)

... Tự làm tiếp nhé

Ta có: \(3x^2+10xy+8y^2=96\)

\(\Leftrightarrow\left(3x^2+6xy\right)+\left(4xy+8y^2\right)=96\)

\(\Leftrightarrow3x\left(x+2y\right)+4y\left(x+2y\right)=96\)

\(\Leftrightarrow\left(3x+4y\right)\left(x+2y\right)=96\) Từ đó ta giải PT nghiệm nguyên ra (Hơi nhiều TH đấy nhé)

Đến phần Ư(96) bạn chỉ cần sử dụng tính chẵn lẻ là sẽ loại bỏ bớt đi 1 số trường hợp rồi

a, \(\left(2x-1\right)^2-36=0\)

\(\Leftrightarrow\left(3x-1-6\right)\left(3x-1+6\right)=0\)

\(\Leftrightarrow\left(3x-7\right)\left(3x+5\right)=0\Leftrightarrow x=\frac{7}{3};-\frac{5}{3}\)

b, \(2x\left(x-3\right)=x^2-9\)

\(\Leftrightarrow2x\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[2x-\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=\pm2\)

c, \(x^2+8x=25\Leftrightarrow x^2+8x-25=0\)

đề sai ko ? 

d, \(\left(3x-2\right)\left(x+6\right)=\left(2-3x\right)\left(3x-5\right)\)

\(\Leftrightarrow\left(3x-2\right)\left(x+6\right)+\left(3x-2\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left[\left(x+6\right)+\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+6+3x-5\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(4x+1\right)=0\Leftrightarrow x=\frac{2}{3};-\frac{1}{4}\)

Ta có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)

Thay vào biểu thức M ta có: \(M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-a\right).\left(-b\right).\left(-c\right)}{abc}=-1\)

TH2: Nếu \(a+b+c\ne0\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)

Thay vào biểu thức M ta có: \(M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{8abc}{abc}=8\)

Vậy \(M=-1\)hoặc \(M=8\)

\(\text{Nếu a+b+c}=0\text{ thì: }M=-1;a+b+c\text{ khác 0 thì:}\)

áp dụng tc dãy tỉ số bằng nhau ta có: a=b=c => M=8

mk lm tóm tắt vì có nhiều câu giống r

có \(f\left(x\right)=\left(x+1\right)A\left(x\right)+5\)

\(f\left(x\right)=\left(x^2+1\right)B\left(x\right)+x+2\)

do f(x) chia cho \(\left(x+1\right)\left(x^2+1\right)\)là bậc 3 nên số dư là bậc 2. ta có \(f\left(x\right)=\left(x+1\right)\left(x^2+1\right)C\left(x\right)+ax^2+bx+c=\left(x+1\right)\left(x^2+1\right)C\left(x\right)+a\left(x^2+1\right)+bx+c-a\)

\(=\left(x^2+1\right)\left(C\left(x\right).x+C\left(x\right)+a\right)+bx+c-a\)

Vậy \(bx+c-a=x+2\Rightarrow\hept{\begin{cases}b=1\\c-a=2\end{cases}}\)

mặt khác ta có \(f\left(-1\right)=5\Leftrightarrow a-b+c=5\Rightarrow a+c=6\Rightarrow\hept{\begin{cases}a=2\\c=4\end{cases}}\)

vậy số dư trong phép chia f(x) cho \(x^3+x^2+x+1\)là \(2x^2+x+4\)