a, thiếu đề 

b, -5 ( 2 - x ) + 4 ( x - 3 ) = 10x - 15 

<=> -10 + 5x + 4x - 12 = 10x - 15 

<=> -x = 7 <=> x = 7 

c,-7(5-x)-2(x-10)=15

<=> -35 + 7x - 2x + 20 = 15 <=> 5x = 30 <=> x = 6 

d, -4(x+1) + 89x - 3 = 24 

<=> 85x = 31 <=> x = 31/85 

e, 5(x-30-2(x+6)) = 9 

<=> 5 (-x-49) = 9 <=> -5x = 254 <=> x = -254/5 

b: =>-10+5x+4x-12=10x-15

=>9x-10x=-15+22

=>-x=7

hay x=-7

c: =>-35+7x-2x+20=15

=>5x-15=15

=>x=6

d: =>-4x-4+72x-24=24

=>68x-32=24

hay x=14/17

\(-7.\left(5-x\right)-2.\left(x-10\right)=1\)

           \(-35+7x-2x+20=1\)

                                 \(-15+5x=1\)

                                                \(5x=1+15\)

                                                \(5x=16\)

                                                   \(x=\frac{16}{5}\)

\(-5.\left(2-x\right)+4.\left(x-3\right)=10x-15\)

        \(-10+5x+4x-12=10x-15\)

                              \(-22+9x=10x-15\)

                                \(9x-10x=-15+22\)

                                            \(-x=7\)

                                                \(x=-7\)

a) \(\dfrac{4n^2}{17n^4}\cdot\dfrac{-7n^2}{12n}\) \(\left(n\ne0\right)\)

\(=\dfrac{4n^2\cdot-7n^2}{17n^4\cdot12n}\)

\(=\dfrac{-28n^4}{204n^5}\)

\(=\dfrac{-7}{51n}\)

b) \(\dfrac{3x-1}{10x^2+2x}\cdot\dfrac{25x^2+10x+1}{1-9x^2}\) \(\left(x\ne\pm\dfrac{1}{3};x\ne0;x\ne-\dfrac{1}{5}\right)\)

\(=\dfrac{3x-1}{2x\left(5x+1\right)}\cdot\dfrac{\left(5x+1\right)^2}{\left(1-3x\right)\left(3x+1\right)}\)

\(=\dfrac{-\left(1-3x\right)\left(5x+1\right)^2}{2x\left(5x+1\right)\left(1-3x\right)\left(1+3x\right)}\) 

\(=\dfrac{-\left(5x+1\right)}{2x\left(1+3x\right)}\)

\(=-\dfrac{5x+1}{6x^2+2x}\)

c) \(\dfrac{27-a^3}{5a+10}:\dfrac{a-3}{3a+6}\) \(\left(a\ne-2;a\ne3\right)\)

\(=\dfrac{\left(3-a\right)\left(9+3a+a^2\right)}{5\left(a+2\right)}\cdot\dfrac{3\left(a+2\right)}{a-3}\)

\(=\dfrac{-\left(a-3\right)\left(a^2+3a+9\right)\cdot3\left(a+2\right)}{5\left(a+2\right)\left(a-3\right)}\)

\(=\dfrac{-3\left(a^2+3x+9\right)}{5}\)

\(=-\dfrac{3x^2+9x+27}{5}\)

d) \(\dfrac{x^2-1}{x^2+2x-15}:\dfrac{x^2+5x+4}{x^2-10x+21}\) \(\left(x\ne3;x\ne-5;x\ne-1;x\ne-4\right)\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}:\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-3\right)\left(x-7\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}\cdot\dfrac{\left(x-3\right)\left(x-7\right)}{\left(x+1\right)\left(x+4\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)

\(x-\frac{3.\left(x+30\right)}{15}-24\frac{1}{2}=\frac{7x}{10}-\frac{2.\left(10x+2\right)}{5}\)

\(\Leftrightarrow\frac{x}{1}-\frac{3x+90}{15}-\frac{49}{2}=\frac{7x}{10}-\frac{20x+4}{5}\)

\(\Leftrightarrow\frac{30x}{30}-\frac{2.\left(3x+90\right)}{15.2}-\frac{49.15}{2.15}=\frac{7x.3}{10.3}-\frac{6.\left(20x+4\right)}{6.5}\)

\(\Leftrightarrow\frac{30x}{30}-\frac{6x+180}{30}-\frac{735}{30}=\frac{21x}{30}-\frac{120x+24}{30}\)

\(\Rightarrow30x-\left(6x+180\right)-735=21x-\left(120x+24\right)\)

\(\Leftrightarrow30x-6x-180-735=21x-120x-24\)

\(\Leftrightarrow24x-915=-99x-24\)

\(\Leftrightarrow24x+99x=-24+915\)

\(\Leftrightarrow123x=891\)

\(\Leftrightarrow x=891:123\)

\(\Leftrightarrow x=\frac{297}{41}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{297}{41}\right\}.\)

Chúc bạn học tốt!

x-\(\frac{3\left(x+30\right)}{15}\)-26=\(\frac{7x}{10}\)-\(\frac{2\left(10x+2\right)}{5}\)

➜\(\frac{30x}{30}\)-\(\frac{6\left(x+30\right)}{30}\)-\(\frac{780}{30}\)=\(\frac{21x}{30}\)-\(\frac{12\left(10x+2\right)}{30}\)

➜ 30x-6x-180-780=21x-120x-24

➜30x-6x-21x-120x=780+180-24

➜Bạn tự tính!!!

a) Ta có: \(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)

\(=x^7-x^6\left(x+1\right)+x^5\left(x+1\right)-...+x\left(x+1\right)+15\)

\(=x^7-x^7-x^6+x^6+x^5-...+x^2+x+15\)

\(=x+15\)

Thay x=79 vào biểu thức \(P\left(x\right)=x+15\), ta được:

\(P\left(79\right)=79+15=94\)

a)x=5

\(\left(3x-15\right)^7=0\Rightarrow3x-15=0\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

\(4^2.x-6=1\Rightarrow16.x=7\)

\(\Rightarrow x=\frac{7}{16}\)

Đề bài câu cuối là gì vậy

Lời giải:
a.

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)

\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)

b)

Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)

\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)