\(\text{Cho }A=\left(\dfrac{3x^2+3}{x^3-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{1}{x-1}\right):\dfrac{2x^2-5x+5}{x-2}\)
\(\text{a, Rút gọn }\)
\(\text{b, Tìm }x\in Z\)\(\text{ để }A\in Z\)
Cho bt: K = \(\left(1-\dfrac{4\sqrt{\text{x}}}{\text{x}-1}+\dfrac{1}{\sqrt{\text{x}}-1}\right):\dfrac{\text{x}-2\sqrt{\text{x}}}{\text{x}-1}\)
a) Rút gọn K
b) Tìm x \(\in\) z để K \(\in\) z
c) Tìm x để K âm
d ) Tìm x để K < \(-\)2
P=\(\left(\dfrac{\text{3x^2+ 3x − 3}}{\text{x^2 + x − 2 }}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right):\dfrac{1}{x^2-1}\)
a,rút gọn P
b,tính P với x2-x-6=0
c,tìm x để p>0
Bạn xem lại \(a,b\) mình làm rồi nha.
\(c,P>0\Leftrightarrow\left(x+1\right)^2>0\) (luôn đúng \(\forall x\))
Vậy với mọi giá trị x thì \(P>0\).
C=\(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{\text{5-x}}{\text{1-x}^{\text{2}}}\right)\):\(\dfrac{1-2x}{\text{x}^{\text{2}}-1}\)
a) Rút gọn
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)
2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
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.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
Câu 1:Cho biểu thức P=\(\text{}\text{}\text{}\text{}\left(\dfrac{x}{4-x^2}+\dfrac{2}{x-2}-\dfrac{1}{x+2}\right):\left(1-\dfrac{x+1}{x+2}\right)\)
a) Rút gọn biểu thức P
b) Tính giá trị của P khi cho \(\left|x\right|\)=1
c)Tìm x để P >0
d)Tìm x để P = \(\dfrac{1}{x+1}\)
Câu 2:Cho tam giác ABC vuông tại A, đường cao AH chia cạnh huyền của tam giác thành hai đoạn có độ dài như sau: HB = 25cm, Hc = 36cm. Vậy đường cao AH có độ dài là
\(\text{Tìm x, biết:}\)
\(a\)) \(20\text{%}x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(b\)) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(c\)) \(\left(x-\dfrac{3}{4}\right)\left(4+3x\right)=0\)
\(d\)) \(x-\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{-26}{5}\)
\(e\)) \(50\text{%}x+\dfrac{2}{3}x=x-5\)
\(g\)) \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}.\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
\(\left(\dfrac{\text{√}x}{\text{√}x+2}+\dfrac{8\text{√}x+8}{x+2\text{√}x}-\dfrac{\text{√}x+2}{\text{√}x}\right):\left(\dfrac{x+\sqrt{x}+3}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
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
Cho bt: K = \(\left(1-\dfrac{4\sqrt{\text{x}}}{\text{x}-1}+\dfrac{1}{\sqrt{\text{x}}-1}\right):\dfrac{\text{x}-2\sqrt{\text{x}}}{\text{x}-1}\)
a) Rút gọn K
b) Tìm x \(\in\) z để K \(\in\) z
c) Tìm x để K âm
d ) Tìm x để K < \(-\)2
a) đkxđ x≥0 , x ≠1
\(K=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)b)
\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\dfrac{1}{\sqrt{x}-2}\)
để K ∈ z thì \(\dfrac{-1}{\sqrt{x}-2}\) nguyên
=> √x -2 ∈ Ư(-1)={-1;1}
=> x ∈ {1; 9}
vậy ...