Giúp gấp vs ạ:
Chứng minh rằng với \(\text{|}x\text{|}\) rất bé so với \(a>0\left(\text{| }x\text{| }\le a\right)\) ta có:
\(\sqrt{a^2+x}\approx a+\dfrac{x}{2a}\left(x>0\right)\)
Áp dụng công thức trên, hãy tính gần đúng số sau:
\(\sqrt{146}\)
Chứng minh rằng với \(\left|x\right|\) rất bé so với \(a>0\left(\left|x\right|\le a\right)\) ta có :
\(\sqrt{a^2+x}\approx a+\dfrac{x}{2a};\left(a>0\right)\)
Áp dụng công thức trên, hãy tính gần đúng các số sau :
a) \(\sqrt{146}\)
b) \(\sqrt{34}\)
c) \(\sqrt{120}\)
B=\(\left(\frac{x\sqrt{x}}{x\text{+}\sqrt{x}\text{+}1}-\frac{1}{x\text{+}\sqrt{x}\text{+}1}\right):\frac{2}{\sqrt{x}\text{+}1}\)
Chứng minh A<0 với mọi 0<x<1
(Đề thi tuyển sinh vào 10 - chuyên Thái Bình)
Cho biểu thức
\(A=\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\) với \(x>0,\text{ }x\ne1\).
a) Rút gọn $A$.
b) Cho \(B=\left(x-\sqrt{x}+1\right).A\) . Chứng minh rằng \(B>1.\)
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )
b) \(B=\left(x-\sqrt{x}+1\right).A\)
Với \(x>0.x\ne1\)thì \(B=\left(x-\sqrt{x}+1\right).\frac{1}{\sqrt{x}}=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}\)
\(B=\left(\sqrt{x}+\frac{1}{\sqrt{x}}-2\frac{\sqrt[4]{x}}{\sqrt[4]{x}}\right)+2\frac{\sqrt[4]{x}}{\sqrt[4]{x}}-1=\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+2-1\)
\(B=\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+1\)
Ta có:
\(\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2\ge0\forall x>0\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}=0\Leftrightarrow\sqrt[4]{x}=\frac{1}{\sqrt[4]{x}}\Leftrightarrow x=1\)(thỏa mãn điều kiện x>0)
Mà theo đề bài, \(x\ne1\)nên dấu bằng không xảy ra
Do đó : \(\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2>0\forall x\left(x>0;x\ne1\right)\)
\(\Rightarrow\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+1>1\forall x\left(x>0;x\ne1\right)\)
\(\Rightarrow B>1\forall x\left(x>0;x\ne1\right)\)
Vậy với \(x>0;x\ne1\)thì \(B>1\)
\(\left(\dfrac{\text{√}x}{\text{√}x+2}+\dfrac{8\text{√}x+8}{x+2\text{√}x}-\dfrac{\text{√}x+2}{\text{√}x}\right):\left(\dfrac{x+\sqrt{x}+3}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
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
Ai giải giúp mấy bài toán vs
Bài 1:
A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)
B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)
Bài 2 rút gọn biểu thức
A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0
B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)
Bài 3 cho biểu thức
P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)
a)Rút gọn P
b)tìm x để P=\(\text{√}x+\frac{5}{2}\)
bài 4 rút gọn biểu thức
A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)
B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)
Bài 5
A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)
a)rút gọn A
b)tìm gtri x để A= -1/4
AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN
Mong mng giúp ạ
câu1 rút gọn
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b)\(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(x\ne\sqrt{2},x\ne-\sqrt{2}\right)\)
c)\(\sqrt{9\text{x}^2}-2\text{x}\left(x< 0\right)\)
d)\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
e)\(\dfrac{x^2-5}{x+\sqrt{5}}\left(x\ne-\sqrt{5}\right)\)
\(a,\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=-1\)
\(b,\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(dk:x\ne\pm\sqrt{2}\right)\\ =\dfrac{x^2+2\sqrt{2}x+\sqrt{2^2}}{x^2-\sqrt{2^2}}\\ =\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\\ =\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\)
\(c,\sqrt{9x^2}-2x\left(dk:x< 0\right)\\ =\sqrt{3^2}.\sqrt{x^2}-2x\\ =3\left|x\right|-2x\\ =-3x-2x\\ =-5x\)
\(d,\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}-3+\sqrt{2}\\ =\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}\\ =\sqrt{2}+3-3+\sqrt{2}\\ =2\sqrt{2}\)
\(e,\dfrac{x^2-5}{x+\sqrt{5}}\left(dk:x\ne-\sqrt{5}\right)\\ =\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\\ =x-\sqrt{5}\)
\(\text{Cho 2 đa thức }f\left(x\right)=x^2-4abx+2a+3\text{ và }g\left(x\right)=x+\sqrt{7-4\sqrt{3}}\)
\(\left(\text{a,b}\in\text{Q }\right).\text{Nếu }f\left(x\right)\text{ chia hết cho }g\left(x\right)\text{ thì giá trị a,b lần lượt là bao nhiêu?}\)
\(g\left(x\right)=0\Leftrightarrow x=-\sqrt{7-4\sqrt{3}}=-\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}-2\)
\(g\left(\sqrt{3}-2\right)=0\Rightarrow f\left(\sqrt{3}-2\right)=0\)
\(\Rightarrow7-4\sqrt{3}-4ab\left(\sqrt{3}-2\right)+2a+3=0\)
\(\Leftrightarrow\sqrt{3}\left(-4-4ab\right)+\left(8ab+2a+10\right)=0\text{ }\left(1\right)\)
Do a, b là các số hữu tỉ nên (1) đúng khi và chỉ khi
\(\int^{-4-4ab=0}_{8ab+2a+10=0}\Leftrightarrow\int^{a=-1}_{b=1}\)
Vậy, \(a=-1;\text{ }b=1.\)
f(x) chia hết cho g(x)
Nếu g(x) =0 hay x = - \(\sqrt{7-4\sqrt{3}}=1-\sqrt{6}\)
=> f( \(1-\sqrt{6}\)) =0
=> \(\left(1-\sqrt{6}\right)^2-4ab\left(1-\sqrt{6}\right)+2a+3=0\)(1)
Cái thứ (2) sử dụng cái gì vậy??? chỉ mình với?
Mình làm sai sao nhiều người tích vậy? Buồn quá!
\(x=-\sqrt{7-4\sqrt{3}}=\sqrt{3}-2\)
\(\left(\sqrt{3}-2\right)^2-4ab\left(\sqrt{3}-2\right)+2a+3=0\)
\(10-4\sqrt{3}-4ab\left(\sqrt{3}-2\right)+2a=0\)
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.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
Cho đa thức : \(\text{A}=\frac{-7}{16}x^3y\cdot\left(2xy^2\right)^3\cdot\left(x^0\right)^2\ \) \(\left(x\ne0\right)\)
a) Thu gọn đơn thức, rồi xác định hệ số, phần biến, bậc của đa thức trên.
b) Biết rằng \(\text{A}<0\). Hãy so sánh giá trị của \(y\) với 0.