Cho\(f\left(x\right)=\frac{1+\sqrt{1+x}}{1+x}+\frac{1-\sqrt[1]{1-x}}{x-1}\) vaf \(a=\frac{\sqrt{3}}{2}\). Tinh \(f\left(a\right)\)
+Tuấn 10B_2 (T ko biết đánh word nên dùng tạm .V)
GPT: \(\(\sqrt{x+3}+\sqrt[3]{x}=3\)\) (Bài này cách lp 9 dễ t ko giải nữa)
Vì \(\(f\left(x\right)=\sqrt{x+3}+\sqrt[3]{x}=3\)\) là hàm tăng trên tập [-3;\(\(+\infty\)\))
Ta có: Nếu \(\(x>1\Leftrightarrow f\left(x\right)>f\left(1\right)=3\)\)nên pt vô nghiệm
Nếu \(\(-3\le x< 1\Leftrightarrow f\left(x\right)< f\left(1\right)=3\)\)nên pt vô nghuêmj
Vậy x = 1
B2, GHPT: \(\(\hept{\begin{cases}2x^2+3=\left(4x^2-2yx^2\right)\sqrt{3-2y}+\frac{4x^2+1}{x}\\\sqrt{2-\sqrt{3-2y}}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\end{cases}}\)\)
ĐK \(\(\hept{\begin{cases}-\frac{1}{2}\le y\le\frac{3}{2}\\x\ne0\\x\ne-\frac{1}{2}\end{cases}}\)\)
Xét pt (1) \(\(\Leftrightarrow2x^2+3-4x-\frac{1}{x}=x^2\left(4-2y\right)\sqrt{3-2y}\)\)
\(\(\Leftrightarrow-\frac{1}{x^3}+\frac{3}{x^2}-\frac{4}{x}+2=\left(4-2y\right)\sqrt{3-2y}\)\)
\(\(\Leftrightarrow\left(-\frac{1}{x}+1\right)^3+\left(-\frac{1}{x}+1\right)=\left(\sqrt{3-2y}\right)^3+\sqrt{3-2y}\)\)
Xét hàm số \(\(f\left(t\right)=t^3+t\)\)trên R có \(\(f'\left(t\right)=3t^2+1>0\forall t\in R\)\)
Suy ra f(t) đồng biến trên R . Nên \(\(f\left(-\frac{1}{x}+1\right)=f\left(\sqrt{3-2y}\right)\Leftrightarrow-\frac{1}{x}+1=\sqrt{3-2y}\)\)
Thay vào (2) \(\(\sqrt{2-\left(1-\frac{1}{x}\right)}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\)\)
\(\(\Leftrightarrow\sqrt{\frac{1}{x}+1}=\frac{\sqrt[3]{x^2\left(x+2\right)}+x+2}{2x+1}\)\)
\(\(\Leftrightarrow\left(2x+1\right)\sqrt{\frac{1}{x}+1}=x+2+\sqrt[3]{x^2\left(x+2\right)}\)\)
\(\(\Leftrightarrow\left(2+\frac{1}{x}\right)\sqrt{1+\frac{1}{x}}=1+\frac{2}{x}+\sqrt[3]{1+\frac{2}{x}}\)\)
\(\(\Leftrightarrow f\left(\sqrt{1+\frac{1}{x}}\right)=f\left(\sqrt[3]{1+\frac{2}{x}}\right)\)\)
\(\(\Leftrightarrow\sqrt{1+\frac{1}{x}}=\sqrt[3]{1+\frac{2}{x}}\)\)
\(\(\Leftrightarrow\left(1+\frac{1}{x}\right)^3=\left(1+\frac{2}{x}\right)^2\)\)
Đặt \(\(\frac{1}{x}=a\)\)
\(\(\Rightarrow Pt:\left(a+1\right)^3=\left(2a+1\right)^2\)\)
Tự làm nốt , mai ra lớp t giảng lại cho ...
Mik ko ngờ bạn lại giải giỏi đến vậy
Mik ko giải được như vậy luôn !!!!
Rút gọn các biểu thức sau
D = \(\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}\)
E =\(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a-1}}\right):\frac{\sqrt{a+1}}{a-2\sqrt{a}+1}\)
F = \(\left(\frac{1}{\sqrt{a-1}}+\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
- Các ĐKXĐ tự tìm dùm mình hen :)
Ta có : \(D=\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}\)
=> \(D=\left(\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{1}{\sqrt{x}+2}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{5+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{1}{\sqrt{x}-3}\right)\left(\sqrt{x}-3\right)=1\)
Ta có : \(E=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{a-2\sqrt{a}+1}\)
=> \(E=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{\left(\sqrt{a}-1\right)^2}\)
=> \(E=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
=> \(E=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
( làm đến đây thôi câu còn lại bạn tự làm hen )
Ghét nhất mấy câu viết sai đề b, c sai rất nhiều bạn ới
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
Cho biểu thức
\(F=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right) \)
a) Rút gọn F
b) Tìm x để F-6<0
Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
\(F=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{2\sqrt{x}+7}{x-4}\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+1\right)\)
a,rút gọn
b,tính F biết x=9-\(4\sqrt{5}\)
c, tìm GTNN của F
\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}+\frac{1}{2-\sqrt{x}}\)
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
d) ĐK: x>=0; x khác 4.
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}.\)
\(D=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(D=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)