a) \(\dfrac{\sqrt{2}}{\sqrt{18}}=\dfrac{1}{3}\)

b) \(\dfrac{\sqrt{15}}{\sqrt{735}}=\dfrac{1}{7}\)

c) \(\dfrac{\sqrt{12500}}{\sqrt{500}}=\dfrac{50\sqrt{5}}{10\sqrt{5}}=\dfrac{50}{10}=5\)

d) \(\dfrac{\sqrt{6^5}}{\sqrt{2^33^5}}=\dfrac{36\sqrt{6}}{18\sqrt{6}}=\dfrac{36}{18}=2\)

\(\sqrt{\dfrac{225}{256}}=\sqrt{\dfrac{15}{16}}=0,9375\)

\(\sqrt{0,0196}=0,14\)

ghi đề kiểu này khó nhìn quá

+) Minor and minimus are miniature repetitions of the daffodil.

+) It was, he held, the final appeal of Ormazd to mankind at large.

+) Dam at Canyon Ferry, on the Missouri river, 18 m.

\(x^2+2xy+y^2-25\)

\(=\left(x+y\right)^2-25\)

\(=\left(x+y-5\right)\left(x+y+5\right)\)

Cái này mà đem đi hỏi thì về quê chăn vịt nhé

a) \(\left(x-1\right)^3=125\)

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

\(\Leftrightarrow x-1=5\)

\(\Leftrightarrow x=5+1\)

\(\Leftrightarrow x=6\)

Vậy \(x=6\)

b) \(2^{x+2}-2^x=96\)

\(\Leftrightarrow\left(2^2-1\right)\cdot2^x=96\)

\(\Leftrightarrow\left(4-1\right)\cdot2^x=96\)

\(\Leftrightarrow3\cdot2^x=96\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

c) \(\left(2x+1\right)^3=343\)

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

\(\Leftrightarrow2x+1=7\)

\(\Leftrightarrow2x=7-1\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

d) \(720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\)

\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\left(đk:x\ne23\right)\)

\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5\)

\(\Leftrightarrow720:\left(46-2x\right)=40\)

\(\Leftrightarrow\dfrac{720}{46-2x}=40\)

\(\Leftrightarrow\dfrac{720}{2\left(23-x\right)}=40\)

\(\Leftrightarrow\dfrac{360}{23-x}=40\)

\(\Leftrightarrow360=40\left(23-x\right)\)

\(\Leftrightarrow9=23-x\)

\(\Leftrightarrow x=23-9\)

\(\Leftrightarrow x=14\left(đk:x\ne23\right)\)

\(\Leftrightarrow x=14\)

Vậy \(x=14\)

e) \(2^x\cdot7=224\)

\(\Leftrightarrow7\cdot2^x=224\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

f) \(\left(3x+5\right)^2=289\)

\(\Leftrightarrow3x+5=\pm17\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=17\\3x+5=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{22}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{22}{3};x_2=4\)

1) \(B=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)+2\left(5-3x\right)^2\)

\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)+2\left(25-30x+9x^2\right)\)

\(=20x^2-20x+5+4x^2+12x-4x-12+50-60+18x^2\)

\(=42x^2-72x+43\)

2) \(C=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a+1\right)^2\)

\(=4a^4-4a^3+2a^2+4a^3-4a^2+2a+2a^2-2a+1-\left(4a^2+4a+1\right)\)

\(=4a^4+2a^2-4a^2+2a^2+1-4a^2-4a-1\)

\(=4a^4-4a^2-4a\)

3) Sky Sơn Tùng làm đúng rồi nhé.

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

\(=x^4+27x^2+1-10x^3+250x^5-1400x^4+1030x^3-302x^2+40x-2\)

\(=-1399x^4-275x^2-1+1020x^3+250x^5+40x\)

5) \(F=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)

\(=\left[a^2+b^2-c^2-\left(a^2-b^2+c^2\right)\right]\cdot\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\)

\(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\cdot2a^2\)

\(=\left(2b^2-2c^2\right)\cdot2a^2\)

\(=2\left(b^2-c^2\right)\cdot2a^2\)

\(=2\left(b-c\right)\left(b+c\right)\cdot2a^2\)

\(=2\cdot2a^2\cdot\left(b-c\right)\left(b+c\right)\)

\(=4a^2\cdot\left(b-c\right)\left(b+c\right)\)

6) \(G=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+\left(-c\right)^2+2ab-2ac-2bc-2\left(a^2+2ab+b^2\right)\)

\(=a^2+b^2+c^2+2ab+a^2+b^2+\left(-c\right)^2+2ab-2a^2-4ab-2b^2\)

\(=0+0+c^2+0+c^2\)

\(=2c^2\)

7) \(H=\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2x\right)\)

\(=a^2-c^2-\left[\left(a-b\right)^2-c^2\right]+b^2-2bx\)

\(=a^2-c^2-\left(a^2-2ab+b^2-c^2\right)+b^2-2bx\)

\(=a^2-b^2-a^2+2ab-b^2+c^2+b^2-2bx\)

\(=2ab-2bx\)

chờ nhé