1 + $\dfrac{\text{1}} {\text{X + 2}}$ = $\dfrac{\text{12}} {\text{8 - x^3}}$
rút gọn các biểu thức sau
\(B=\dfrac{3\text{x}^2+6\text{x}+12}{x^3-8\dfrac{ }{ }}\)
C=\(\left(\dfrac{x+1}{2\text{x}-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2\text{x}+2}\right).\dfrac{4\text{x}^2-4}{5}\)
E=\(\dfrac{x^2-10\text{x}+25}{x^2-5\text{x}}\)
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
Giải pt sau
\(\dfrac{1}{\text{x}^2+5x+6}\)+\(\dfrac{1}{\text{x}^2+7x+12}\)+\(\dfrac{1}{x^2+9x +12}\)+\(\dfrac{1}{\text{x}^2+9x+30}=\dfrac{1}{8}\)
Sửa lại đề nha:
\(\dfrac{1}{x^2+9x+12}thành\dfrac{1}{x^2+9x+20}\)
⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)
⇔ \(x^2+8x+12=32\)
⇔ \(x^2+8x-20=0\)
⇔ \(\left(x-2\right)\left(x+10\right)=0\)
⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
1. Rút gọn biểu thức A = \(\dfrac{\text{√ x + 1}}{\text{√ x − 1 }}-\dfrac{\text{√ x − 1}}{\text{√ x + 1}}+\dfrac{\text{8 √ x}}{\text{1 − x }}\)
2. Rút gọn biểu thức B = \(\dfrac{\text{√ x − x − 3}}{\text{x − 1 }}-\dfrac{\text{1}}{\text{√ x − 1 }}\) với x ≥ 0, x ≠ 1
\(\left(\dfrac{\text{√}x}{\text{√}x+2}+\dfrac{8\text{√}x+8}{x+2\text{√}x}-\dfrac{\text{√}x+2}{\text{√}x}\right):\left(\dfrac{x+\sqrt{x}+3}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
a) rút gọn P
b)CMR: P≤1
b) (4√x + 4)/(x + 2√x + 5) ≥ 1
⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
Do x ≥ 0 ⇒ x + 2√x + 5 > 0
⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0
⇔ 4√x + 4 - x - 2√x - 5 ≤ 0
⇔ -x + 2√x - 1 ≤ 0
⇔ -(x - 2√x + 1) ≤ 0
⇔ -(√x - 1)² ≤ 0 (luôn đúng)
Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ≥ 0
a: \(P=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+5}\)
b: 4(căn x+1)>=4
x+2căn x+5>=5
=>P<=4/5<1
Tìm số thập phân x biết:
a) \(^{\text{x}}\)\(^{ }\)+ \(\dfrac{3}{4}\)= \(\dfrac{4}{5}\) | b) \(^{\text{x}}\) - \(\dfrac{1}{2}\) = \(\dfrac{5}{8}\) | c) \(^{\text{x}}\) x \(\dfrac{5}{6}\) = \(\dfrac{4}{5}\) | d)\(^{\text{x}}\) : \(\dfrac{5}{8}\) = \(\dfrac{1}{25}\) |
a) \(x=0,05\)
b) \(x=1,125\)
c) \(x=0,96\)
d) \(x=0,025\)
Bạn tự làm đi dễ mà . Cố mag vận động đầu óc đừng copy làm bài nữa khó lắm mới hỏi thôi
1 a..Rút gọn biểu thức A = \(\dfrac{\text{ x 2 − 4 x + 4}}{\text{x 3 − 2 x 2 − ( 4 x − 8 ) }}\)
b. Rút gọn biểu thức B = \(\left(\dfrac{x+2}{\text{x }\sqrt{\text{x }}+1}-\dfrac{1}{\sqrt{\text{x}}+1}\right).\dfrac{\text{4 }\sqrt{x}}{3}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
a,\(\dfrac{1}{7}\text{x}\dfrac{2}{7}+\dfrac{1}{7}\text{x}\dfrac{5}{7}+\dfrac{6}{7}\) b,\(\dfrac{6}{11}\text{x}\dfrac{4}{9}+\dfrac{6}{11}\text{x}\dfrac{7}{9}-\dfrac{6}{11}\text{x}\dfrac{2}{9}\)
c, \(\dfrac{4}{25}\text{x}\dfrac{5}{8}\text{x}\dfrac{25}{4}\text{x}24\)
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)
\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)
c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)
\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)
\(=1.15=15\)
Giải các bất phương trình sau:
1) \(\dfrac{\text{x}-1}{x-3}>1\) 2) \(\sqrt{\text{x}^2+x-12}< 8-x\)
1:
ĐKXĐ: x<>3
\(\dfrac{x-1}{x-3}>1\)
=>\(\dfrac{x-1-\left(x-3\right)}{x-3}>0\)
=>\(\dfrac{x-1-x+3}{x-3}>0\)
=>\(\dfrac{2}{x-3}>0\)
=>x-3>0
=>x>3
2: ĐKXĐ: \(\left[{}\begin{matrix}x>=3\\x< =-4\end{matrix}\right.\)
\(\sqrt{x^2+x-12}< 8-x\)
=>\(\left\{{}\begin{matrix}8-x>=0\\x^2+x-12< \left(8-x\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =8\\x^2+x-12-x^2+16x-64< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =8\\17x-76< 0\end{matrix}\right.\)
=>\(x< \dfrac{76}{17}\)
Kết hợp ĐKXĐ, ta được: \(\left[{}\begin{matrix}3< =x< \dfrac{76}{17}\\x< =-4\end{matrix}\right.\)
Cho x + 3y - 2z = 36. Tìm x,y,z biết
a) \(\dfrac{\text{x-1}}{\text{3}}=\dfrac{\text{y+2}}{\text{4}}=\dfrac{\text{z-2}}{\text{3}}\)
b) \(\dfrac{\text{x}}{\text{4}}=\dfrac{\text{y}}{\text{3}};\dfrac{\text{y}}{\text{2}}=\dfrac{\text{z}}{\text{5}}\)
c) 9x = 5y ; 2x = z
d) 2x = 3y = 4z
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9