\(\frac{7}{3}\times\frac{21}{13}-\frac{7}{3}\times\frac{8}{13}=\frac{7}{3}\times\left(\frac{21}{13}-\frac{8}{13}\right)\)

\(=\frac{7}{3}\times\frac{13}{13}=\frac{7}{3}\times1=\frac{7}{3}\)

\(20)\ 5^{36} = (5^3)^{12} = 125^{12}\)

\(11^{24} = (11^2)^{12} = 121^{12}\)

vì 125 > 121 nên \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

\(21)\ 2^{225} = (2^3)^{75} = 8^{75}\)

\(3^{150} = (3^2)^{75} = 9^{75}\)

vì 8 < 9 nên \(8^{75}<9^{75}\Rightarrow2^{225}<3^{150}\)

\(22)\ 3^{4000} = (3^2)^{2000} = 9^{2000}\)

vì 9 > 2 nên \(9^{2000}>2^{2000}\Rightarrow3^{4000}>2^{2000}\)

\(23)\ 2^{333} = (2^3)^{111} = 8^{111}\)

\(3^{222} = (3^2)^{111} = 9^{111}\)

vì 8 < 9 nên \(8^{111}<9^{111}\Rightarrow2^{333}<3^{222}\)

\(24)\ 99<9999\Rightarrow99^{10}<9999^{10}\)

\(25)\ 3^{12} = (3^3)^4 = 27^4\)

\(5^8 = (5^2)^4 = 25^4\)

vì 27>25 nên \(27^4>25^4=>3^{12}>5^8\)

\(26)\ 8^{12} = (2^3)^{12} = 2^{36} = (2^9)^4 = 512^4\)

\(12^8 = (12^2)^4 = 144^4\)

vì 512 > 144 nên \(512^4>144^4\Rightarrow8^{12}>12^8\)

\(27)\ 3^{2444} = 3^{22 \times 111 + 2} = (3^{22})^{111} \times 9\)

\(4^{333} = (4^3)^{111} = 64^{111}\)

vì \(3^{22} \times 9 > 64\) nên \((3^{22})^{111}\times9>64^{111}\Rightarrow3^{2444}>4^{333}\)

\(28)\ 9^{12} = (3^2)^{12} = 3^{24}\)

\(27^7 = (3^3)^7 = 3^{21}\)

vì 24 > 21 nên \(3^{24}>3^{21}\Rightarrow9^{12}>27^7\)

\(29)\ 27^{11} = (3^3)^{11} = 3^{33}\)

\(81^8 = (3^4)^8 = 3^{32}\)

vì 33>32 nên \(3^{33}>3^{32}\Rightarrow27^{11}>81^8\)

\(30)\ 64^8 = (2^6)^8 = 2^{48}\)

\(16^2 = (2^4)^2 = 2^8\)

vì 48 > 8 ênn \(2^{48}>2^8\Rightarrow64^8>16^2\)

\(31)\ 2^{300} = (2^3)^{100} = 8^{100}\)

\(5^{200} = (5^2)^{100} = 25^{100}\)

vì 8 < 25 nên : \(8^{100}<25^{100}\Rightarrow2^{300}<5^{200}\)

\(32)\ 3^{200} = (3^2)^{100} = 9^{100}\)

\(2^{300} = (2^3)^{100} = 8^{100}\)

vì 9>8 nên \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

\(33)\ 10^{30} = (10^3)^{10} = 1000^{10}\)

\(2^{100} = (2^{10})^{10} = 1024^{10}\)

vì 1000<1024 nên \(1000^{10}<1024^{10}\Rightarrow10^{30}<2^{100}\)

\(34)\ 5^{30} = (5^3)^{10} = 125^{10}\)

vì 125>124 nên \(125^{10}>124^{10}\Rightarrow5^{30}>124^{10}\)

dạng 1

bài 2: đkxđ: \(x \ge 0; x \neq 1\)

a. thay x = 25 vào biểu thức A ta được:

\(A=\frac{\sqrt{25}-2}{\sqrt{25}+1}=\frac{5-2}{5+1}=\frac36=\frac12\)

b. \(B = \frac{1}{\sqrt{x}+1} - \frac{\sqrt{x}}{1-\sqrt{x}} + \frac{2}{x-1}\)

\(B = \frac{1}{\sqrt{x}+1} + \frac{\sqrt{x}}{\sqrt{x}-1} + \frac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(B = \frac{\sqrt{x}-1 + \sqrt{x}(\sqrt{x}+1) + 2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(B = \frac{\sqrt{x}-1 + x + \sqrt{x} + 2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(B = \frac{x + 2\sqrt{x} + 1}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(B = \frac{(\sqrt{x}+1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(B = \frac{\sqrt{x}+1}{\sqrt{x}-1} \quad (\text{đpcm})\)

\(c.P=A:B\)

\(P = \frac{\sqrt{x}-2}{\sqrt{x}+1} : \frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(P = \frac{\sqrt{x}-2}{\sqrt{x}+1} \cdot \frac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(P = \frac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}+1)^2}\)

vì P = 5/2 nên: \(\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}+1)^2} = \frac{5}{2}\)

\(\Rightarrow2(\sqrt{x}-2)(\sqrt{x}-1)=5(\sqrt{x}+1)^2\)

\(\Rightarrow2(x-3\sqrt{x}+2)=5(x+2\sqrt{x}+1)\)

\(\Rightarrow2x-6\sqrt{x}+4=5x+10\sqrt{x}+5\)

\(\Rightarrow3x+16\sqrt{x}+1=0\)

vì \(x \ge 0 \Rightarrow \sqrt{x} \ge 0\) nên \(3x+16\sqrt{x}+1\ge1>0\quad\left(\forall x\ge0\right)\)

⇒ phương trình trên vô nghiệm

vậy không có giá trị x thoả P = 5/2

bài 3: đkxđ: \(x \ge 0; x \neq 1; x \neq 4\)

thay x = 9 vào biểu thức A ta được: \(A=\frac{4\sqrt{9}}{\sqrt{9}-1}=\frac{4 \cdot3}{3 - 1}=\frac{12}{2}=6\)

vậy khi x = 9 thì A = 6

\(B = \frac{\sqrt{x}+5}{\sqrt{x}+2} - \frac{1}{2-\sqrt{x}} + \frac{12}{x-4}\)

\(B = \frac{\sqrt{x}+5}{\sqrt{x}+2} + \frac{1}{\sqrt{x}-2} + \frac{12}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

\(B = \frac{(\sqrt{x}+5)(\sqrt{x}-2) + 1 \cdot (\sqrt{x}+2) + 12}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

\(B = \frac{x - 2\sqrt{x} + 5\sqrt{x} - 10 + \sqrt{x} + 2 + 12}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

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

\(B = \frac{(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

\(B = \frac{\sqrt{x}+2}{\sqrt{x}-2}\)

c. ta có \(P = A \cdot B\)

\(\Rightarrow P=\frac{4\sqrt{x}}{\sqrt{x}-1}\cdot\frac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(\Rightarrow P=\frac{4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}-2)}\)

vì P = 2 nên \(\frac{4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}-2)} = 2\)

\(\Rightarrow2\sqrt{x}(\sqrt{x}+2)=(\sqrt{x}-1)(\sqrt{x}-2)\)

\(\Rightarrow2x+4\sqrt{x}=x-3\sqrt{x}+2\)

\(\Rightarrow x+7\sqrt{x}-2=0\)

đặt \(t = \sqrt{x}\) với \(t\ge0,t\neq1,t\neq2\)

ta có: \(t^2 + 7t - 2 = 0\)

\(\Delta = 7^2 - 4 \cdot 1 \cdot (-2) = 49 + 8 = 57 > 0\)

\(\Rightarrow\begin{cases}t_1=\frac{-7 + \sqrt{57}}{2}\left(N\right)\\ t_2=\frac{-7 - \sqrt{57}}{2}<0\left(L\right)\end{cases}\)

\(x=t^2=\left(\frac{\sqrt{57}-7}{2}\right)^2=\frac{57 - 14\sqrt{57} + 49}{4}\)

\(=\frac{106 - 14\sqrt{57}}{4}=\frac{53 - 7\sqrt{57}}{2}\)

\(400\,000 + 60\,000 - 40\,000\)

\(=460\,000-40\,000=420\,000\)

\(590\,000 - (50\,000 + 40\,000)\)

\(=590\,000-90\,000=500\,000\)

\(a.\frac{x-3}{x+7}=\frac{1}{6}\) (đkxđ: \(x+7\neq0\Rightarrow x\neq-7\) )

\(\Rightarrow 6(x - 3) = 1(x + 7)\)

\(\Leftrightarrow 6x - 18 = x + 7\)

\(\Leftrightarrow 6x - x = 7 + 18\)

\(\Leftrightarrow5x=25\Rightarrow x=5\left(TM\right)\)

b. \(\frac{x+4}{x-3}=\frac{x-5}{x+1}\) (đkxđ: \(\begin{cases} x - 3 \neq 0 \\ x + 1 \neq 0 \end{cases} \Leftrightarrow \begin{cases} x \neq 3 \\ x \neq -1 \end{cases}\) )

\(\Rightarrow (x + 4)(x + 1) = (x - 5)(x - 3)\)

\(\Leftrightarrow x^2 + x + 4x + 4 = x^2 - 3x - 5x + 15\)

\(\Leftrightarrow x^2 + 5x + 4 = x^2 - 8x + 15\)

\(\Leftrightarrow x^2 + 5x - x^2 + 8x = 15 - 4\)

\(\Leftrightarrow13x=11\Rightarrow x=\frac{11}{13}\left(TM\right)\)

bài 1:

1. \(9x-15y=3(3x-5y)\)

2. \(8x^2+12x-4=4(2x^2+3x-1)\)

3. \(-5x^2-25xy+10y^2=-5(x^2+5xy-2y^2)\)

4. \(x^3-x^2y+xy^2=x(x^2-xy+y^2)\)

5. \(x^3-2x^2+2x-1=(x^3-1)-(2x^2-2x)\)

\(= (x - 1)(x^2 + x + 1) - 2x(x - 1)\)

\(= (x - 1)(x^2 + x + 1 - 2x)\)

\(= (x - 1)(x^2 - x + 1)\)

6. \(xy + 3x - 7y - 21\)

\(= (xy + 3x) - (7y + 21)\)

\(= x(y + 3) - 7(y + 3)\)

\(= (y + 3)(x - 7)\)

7. \(x^2-2x-4y^2-4y=(x^2-4y^2)-(2x+4y)\)

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

8. \(x^4+x^3-4x-4=(x^4+x^3)-(4x+4)\)

\(=x^3(x+1)-4(x+1)=(x+1)(x^3-4)\)

9. \(x^2-y^2-2x-2y=(x^2-y^2)-(2x+2y)\)

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

10. \(64x^3+1=(4x)^3+1^3\)

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

\(= (4x + 1)(16x^2 - 4x + 1)\)

bài 2:

\(A = x^2 + xy - 5x - 5y\)

\(A = (x^2 + xy) - (5x + 5y)\)

\(A = x(x + y) - 5(x + y)\)

\(A = (x + y)(x - 5)\)

thay \(x = 15\frac{1}{5}\) và \(y = 14\frac{4}{5}\) vào A ta được:

\(A = 30 \cdot \frac{51}{5} = 6 \cdot 51 = 306\)

\(B = 3xy - 8y - 15x + 40\)

\(B = (3xy - 15x) - (8y - 40)\)

\(B = 3x(y - 5) - 8(y - 5)\)

\(B = (y - 5)(3x - 8)\)

thay x = 1999, y = 5 vào B ta được:

(5 - 5)(3*1999-8) = 0

\(C = y^3 + 4x^2y + 4xy + 8x^3 + 2xy^2\)

\(C = (8x^3 + y^3) + (4x^2y + 2xy^2) + 4xy\)

\(C = (2x + y)[(2x)^2 - 2x \cdot y + y^2] + 2xy(2x + y) + 4xy\)

\(C = (2x + y)(4x^2 - 2xy + y^2) + 2xy(2x + y) + 4xy\)

thay 2x+y=1 vào C ta được:

\(C = 1 \cdot (4x^2 - 2xy + y^2) + 2xy \cdot 1 + 4xy\)

\(C = 4x^2 - 2xy + y^2 + 2xy + 4xy\)

\(C = 4x^2 + 4xy + y^2\)

\(C=(2x+y)^2=1^2=1\)

bài 3:

a. \(x^2 - 7x + 12 = x^2 - 3x - 4x + 12\)

\(=(x^2-3x)-(4x-12)=x(x-3)-4(x-3)\)

\(= (x - 3)(x - 4)\)

b. \(x^2 - 11x + 30 = x^2 - 5x - 6x + 30\)

\(= (x^2 - 5x) - (6x - 30)\)

\(=x(x-5)-6(x-5)=(x-5)(x-6)\)

c. \(x^4 + 4y^4 = (x^2)^2 + (2y^2)^2 + 4x^2y^2 - 4x^2y^2\)

\(= [(x^2)^2 + 2 \cdot x^2 \cdot 2y^2 + (2y^2)^2] - (2xy)^2\)

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

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

d. \(4x^4y^4 + 1 = (2x^2y^2)^2 + 1^2 + 4x^2y^2 - 4x^2y^2\)

\(= [(2x^2y^2)^2 + 2 \cdot 2x^2y^2 \cdot 1 + 1^2] - (2xy)^2\)

\(= (2x^2y^2 + 1)^2 - (2xy)^2\)

\(= (2x^2y^2 + 1 - 2xy)(2x^2y^2 + 1 + 2xy)\)

What was the weather like there?

five