Chúng ta sẽ sử dụng hằng đẳng thức em nhé :)

a. \(x^3+1-x^3+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b. \(x^2+x-6+3x=4x+3\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

c. \(x^3+2x^2-5x-10+5x=2x^2+17\Leftrightarrow x^3=27\Leftrightarrow x=3\)

a,

(x²-x+1)(x+1)-x³+3x=15

x³-x²+x+x²-x+1-x³+3x=15

x³-x³-x²+x²+x-x+3x+1=15

3x+1=15

3x=15-1

3x=14

x=14/3

b,

(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)

Tới đây hết biết , đề có gì sai sai sao ý !

c,

(x²-5)(x+2)+5x=2x²+17

x³+2x²-5x-10+5x=2x²+17

x³+2x²-5x+5x-10=2x²+17

x³+2x²-10=2x²+17

x³-10=17

x³=17+10

x³=27

\(\Rightarrow x=3\)(Vì : 3³=27)

_k_ nhé bn

Nhân ra thôi bạn, có hằng đẳng thức gì đâu !

a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

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

\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)

\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)

\(\Leftrightarrow x^2+4x-6=4x+3\)

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)

\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)

1)\(x=\left(\dfrac{1}{5}+\dfrac{2}{3}\right):\dfrac{4}{7}=\left(\dfrac{3}{15}+\dfrac{10}{15}\right):\dfrac{4}{7}=\dfrac{13}{15}:\dfrac{4}{7}=\dfrac{91}{60}\)

2)\(x=\left(-\dfrac{3}{10}-\dfrac{3}{4}\right):\dfrac{1}{2}=\left(-\dfrac{6}{20}-\dfrac{15}{20}\right):\dfrac{1}{2}=-\dfrac{21}{20}:\dfrac{1}{2}=-\dfrac{21}{10}\)

3)\(x=-\dfrac{24}{36}-\dfrac{7}{15}=-\dfrac{2}{3}-\dfrac{7}{15}=\dfrac{-10}{15}-\dfrac{7}{15}=-\dfrac{17}{15}\)

1) \(\dfrac{4}{7}x=\dfrac{1}{5}+\dfrac{2}{3}\)

\(\dfrac{4}{7}x=\dfrac{13}{15}\)

\(x=\dfrac{91}{60}\)

2) \(\dfrac{1}{2}x=-\dfrac{3}{10}-\dfrac{3}{4}\)

\(\dfrac{1}{2}x=-\dfrac{21}{20}\)

\(x=-\dfrac{21}{10}\)

3) \(x=-\dfrac{24}{36}-\dfrac{7}{15}\)

\(x=\dfrac{-2}{3}+\dfrac{7}{15}\)

\(x=-\dfrac{17}{15}\)

1) 4/7 x - 2/3 = 1/5

4/7 x = 1/5 + 2/3

4/7 x = 13/15

x = 13/15 : 4/7

x = 91/60

2)1/2 x + 3/4 = -3/10

1/2 x = -3/10 - 3/4 

1/2 x = -21/20

x = -21/20 : 1/2

x = -21/10

3)x+7/15 = -24/36

x+7/15 = -2/3

x = -2/3 - 7/15

x = -17/15.

a) B = \(\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{21}{20}=\dfrac{1}{3}.1.....\dfrac{21}{1}=\dfrac{21}{3}=7\)

b) Em chịu, chưa học số âm :)

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

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

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

2. a) x + (-23) = (-100) + 77

1)

a: (36-16).(-5)+6.(-14-6)

=20.(-5)+6.(-20)

=20.(-5)+(-6).20

=20.(-5+-6)

=20.(-11)

=-220

2)

a/x+(-23)=(-100)+77

x+(-23)=-23

x=(-23)+(-23)

x=0

Vậy x=0

Giải pt à bạn

Bài 1:

a) \(x\left(x+1\right)+x\left(x-1\right)-2x^2\)

\(=x^2+x+x^2-x-2x^2\)

\(=2x^2-2x^2\)

\(=0\)

b) \(\left(x+2\right)\left(x^2-x+1\right)-\left(x-2\right)\left(x^2+x+1\right)\)

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

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

\(=2x^2+4\)

c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)

\(=\left(x-3+x+7\right)^2\)

\(=\left(2x+4\right)^2\)

Bài 2:

a) \(x^2-4-4\left(2-x\right)\)

\(=\left(x^2-4\right)+4\left(x-2\right)\)

\(=\left(x^2-2^2\right)+4\left(x-2\right)\)

\(=\left(x+2\right)\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x+2+4\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

b) \(x^3+8\left(1-2x\right)^3\)

\(=x^3+\left[2\left(1-2x\right)\right]^3\)

\(=\left[x+2\left(1-2x\right)\right]\left[x^2-x\cdot2\left(1-2x\right)+2^2\left(1-2x\right)^2\right]\)

\(=\left(x+2-4x\right)\left[x^2-2x\left(1-2x\right)+4\left(1-4x+4x^2\right)\right]\)

\(=\left(2-3x\right)\left(x^2-2x+4x^2+4-16x+16x^2\right)\)

\(=\left(2-3x\right)\left(21x^2-18x+4\right)\)

a, ĐKXĐ:\(x\ne-3\)

\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ:\(x>2\)

\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)