Bài \(1\)

\(a)\)\(\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\)

\(b)\)\(\left(x+2\right)^3=x^3+3.x^2.2+3.x.2^2+2^3=x^3+6x^2+12x+8\)

\(c)\)\(\left(x+3\right)^3=x^3+3.x^2.3+3.x.3^2+3^3=x^3+9x^2+27x+27\)

\(d)\)\(\left(x+4\right)^3=x^3+3.x^2.4+3.x.4^2+4^3=x^3+12x^2+48x+64\)

\(e)\)\(\left(a+5\right)^3=a^3+3.a^2.5+3.a.5^2+5^3=a^3+15a^2+75a+125\)

\(f)\)\(\left(10+x\right)^3\)

\(=10^3+3.10^2.x+3.10.x^2+x^3\)

\(=1000+300x+30x^2+x^3\)

\(=x^3+30x^2+300x+1000\)

\(g)\)\(\left(2x+1\right)^3\)

\(=\left[\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\right]\)

\(=8x^3+12x^2+6x+1\)

\(h)\)\(\left(1+3x\right)^3\)

\(=1^3+3.1^2.3x+3.1.\left(3x\right)^2+\left(3x\right)^3\)

\(=1+9x+27x^2+27x^3\)

\(=27x^3+27x^2+9x+1\)

\(i)\)\(\left(3x+\dfrac{1}{2}\right)^3\)

\(=\left(3x\right)^3+3.\left(3x\right)^2.\dfrac{1}{2}+3.3x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

\(=27x^3+\dfrac{27}{2}x^2+\dfrac{9}{4}x+\dfrac{1}{8}\)

\(j)\)\(\left(2x+\dfrac{1}{3}\right)^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{3}+3.2x.\left(\dfrac{1}{3}\right)^2+ \left(\dfrac{1}{3}\right)^3\)

\(=8x^3+4x^2+\dfrac{2}{3}x+\dfrac{1}{27}\)

Bài \(2\)

\(a)\) \(2.2.2.2.2.2.2.2=2^8\)

\(b)\) \(10.10.10.10.10=10^5\)

\(c)\) \(8.8.8.6.6.7.7.7=8^3.6^2.7^3\)

\(d)\) \(a.a.a.a.a.a.a.a.a=a^9\)

\(e)\) \(10000 . 10 . 10 . 10 . 100\)

\(=10^4.10.10.10.10^2=10^9\)

\(f)\) \(2x.2x.2x.2x.2x=\left(2x\right)^5\)

\(95-\left(3x-27\right).39=4212\)

\(95-\left(3x-27\right)=4212:39\)

\(95-\left(3x-27\right)=108\)

\(3x-27=95-108\)

\(3x-27=-13\)

\(3x=-13+27\)

\(3x=14\)

\(x=14:3\)

\(x=\dfrac{14}{3}\)

Vậy \(x=\dfrac{14}{3}\)

\(3.\left(\dfrac{2}{3}-7x\right)=\dfrac{5}{3}-12x\)

\(\Rightarrow2-21=\dfrac{5}{3}-12x\)

\(\Rightarrow-21+12x=\dfrac{5}{3}-2\)

\(\Rightarrow-9x=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}:\left(-9\right)\)

\(\Rightarrow x=\dfrac{-1}{3}.\dfrac{-1}{9}\)

\(\Rightarrow x=\dfrac{1}{27}\)

Vậy \(x=\dfrac{1}{27}\)

__________

\(\rightarrow\) Mình nghĩ đầu bài có vấn đề lớp 6 mà sao bài khó thế ạ nên mình đã sửa lại đề ạ.

\(\dfrac{5}{9}:\dfrac{11}{24}=\dfrac{5}{9}.\dfrac{24}{11}=\dfrac{5.24}{9.11}=\dfrac{5.3.8}{3.3.11}=\dfrac{40}{33}>1\)

Vậy \(\Rightarrow\dfrac{5}{9}>\dfrac{11}{24}\) 

cậu ơi hình như cậu lm sai đề r á.

\(D=\left(2x+1\right)^2-\left(2x-3\right)^2+6x\)

\(D=\left[\left(2x\right)^2+2.2x.1+1^2\right]-\left[\left(2x\right)^2-2.2x.3+3^2\right]+6x\)

\(D=4x^2+4x+1-\left(4x^2-12x+9\right)+6x\)

\(D=4x^2+4x+1-4x^2+12x-9+6x\)

\(D=22x-8\)

___________

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

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

\(E=x^2-8x+16-x^2-2x-2x+3\)

\(E=-12x+19\)

Đánh lẽ phải bỏ dấu ngoặc và đổi dấu chứ nhỉ??

Bài \(1\)

\(1)\) \(1-5+7-8+4-1+5-7+8\)

\(=(1-1)+(5-5)+(7-7)+(8-8)\)

\(=0+0+0+0\)

\(=0\)

\(2)\) \(14-23+(5+14)-(5-23)+17\)

\(=14-23+5+14-5+23+17\)

\(=(14+14)+(23-23)+(5-5)+17\)

\(=28+17\)

\(=45\)

\(3)\) \(12-44+9-3+14-19-9-12\)

\(=(12-12)+(9-9)+(14-44)+3\)

\(=-30+3\)

\(=-33\)

\(4)\) \(22-(4-8+12)+(-8-12+4)\)

\(=22-4+8-12-8-12+4\)

\(=22+(4+4)+(8-8)+(-12-12)\)

\(=22-24\)

\(=-2\)

\(a)\)

\(210:x-\dfrac{1}{2}=20,5\)

\(210:x=20,5+\dfrac{1}{2}\)

\(210:x=21\)

\(x=210:21\)

\(x=10\)

Vậy \(x=10\)

\(b)\)

\(7.3^x+20.3^x=3^{25}\)

\(3^x.\left(7+20\right)=3^{25}\)

\(3^x.27=3^{25}\)

\(3^x.3^3=3^{25}\)

\(3^x=3^{25}:3^3\)

\(3^x=3^{22}\)

\(x=22\)

Vậy \(x=22\)