Cho a,b,c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\);\(a+b+c=23\);\(\sqrt{abc}=3\)
Tính giá trị biếu thức A=\(\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ca}+\sqrt{b}-6}\)
cho a, b, c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\)
tính giá trị biểu thức \(A=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath
Em tham khảo nhé!
cho a,b,c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\)Tính giá trị biểu thức\(N=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ac}+\sqrt{b}-6}\)
Từ giả thiết: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}\)
Xét hạng tử: \(\frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}\)
Từ đó: \(N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}\)
\(=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Mặt khác: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13\)
Suy ra: \(N=\frac{4}{9-13}=-1\). Kết luận: N = -1.
Từ giả thiết: \sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}a+b+c=7⇔c=7−a−b
Xét hạng tử: \frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}ab+c−61=ab+7−a−b−61=(a−1)(b−1)1
Từ đó: N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}N=(a−1)(b−1)1+(b−1)(c−1)1+(c−1)(a−1)1
=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}=(a−1)(b−1)(c−1)a+b+c−3=abc−(ab+bc+ca)+(a+b+c)−1a+b+c−3
=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=3−(ab+bc+ca)+7−17−3=9−(ab+bc+ca)4
Mặt khác: \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13ab+bc+ca=2(a+b+c)2−(a+b+c)=13
Suy ra: N=\frac{4}{9-13}=-1N=9−134=−1. Kết luận: N = -1.
Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
mà \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23\)nên \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=13\)
Ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Rightarrow\sqrt{c}-6=-\sqrt{a}-\sqrt{b}+1\)
nên \(\sqrt{ab}+\sqrt{c}-6=\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
Tương tự \(\hept{\begin{cases}\sqrt{bc}+\sqrt{a}-6=\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)\\\sqrt{ac}+\sqrt{b}-6=\left(\sqrt{a}-1\right)\left(\sqrt{c}-1\right)\end{cases}}\)
Vậy \(H=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ac}+\sqrt{b}-6}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{c}-1+\sqrt{a}-1+\sqrt{b}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-3}{\sqrt{abc}+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=\frac{7-3}{3+7-13-1}=-1\)
Cho a, b, c thỏa mãn: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\). Tính giá trị của biểu thức: \(H=\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Cho các số a,b,c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7,a+b+c=23,\sqrt{abc}=3\)
Tính GTBT:
H=\(\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ac}+\sqrt{b}-6}\)
Cho a,b,c là 3 số dương thỏa mãn: \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=1\)Tính giá trị của biểu thức M =\(\frac{\sqrt{a}-\sqrt{b}}{c+\sqrt{abc}}+\frac{\sqrt{b}-\sqrt{c}}{a+\sqrt{abc}}+\frac{\sqrt{c}-\sqrt{a}}{b+\sqrt{abc}}\)
Cho các số thực dương a,b,c thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\). CMR:
\(\frac{a+b}{\sqrt{ab+c}}+\frac{b+c}{\sqrt{bc+a}}+\frac{c+a}{\sqrt{ca+b}}\ge3\sqrt[6]{abc}\)
Giải:
\(GT\Leftrightarrow ab+bc+ca\ge abc\)
\(\Rightarrow ab\le\frac{ab+bc+ca}{c}\)
\(\Rightarrow\frac{a+b}{\sqrt{ab+c}}\ge\frac{a+b}{\sqrt{\frac{ab+bc+ca}{c}+c}}=\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Tương tự rồi cộng lại: \(VT\ge\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}+\frac{\left(b+c\right)\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\left(c+a\right)\sqrt{c}}{\sqrt{\left(b+a\right)\left(b+c\right)}}\)\(\ge3\sqrt[3]{\sqrt{abc}}=3\sqrt[6]{abc}\)
Lần sau mấy bạn hỏi bài thì đăng lên nhé!
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
OMG !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Cho a,b,c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\) ; \(a+b+c=23\) ; \(\sqrt{abc}=3\)
Tính giá trị biểu thức A=\(\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\Leftrightarrow7^2=23+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=13\)
Ta lại có \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}-6=-\sqrt{a}-\sqrt{b}+1\Leftrightarrow\sqrt{ab}+\sqrt{c}-6=\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
Chứng minh tương tự:
\(\sqrt{bc}+\sqrt{a}-6=\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)\)
\(\sqrt{ac}+\sqrt{b}-6=\left(\sqrt{a}-1\right)\left(\sqrt{c}-1\right)\)
Vậy A=\(\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}=\dfrac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\dfrac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\dfrac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{c}-1+\sqrt{a}-1+\sqrt{b}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-3}{\sqrt{abc}+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}=\dfrac{7-3}{3+7-13-1}=-1\)
1: Cho biểu thức \(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
a: Tìm ĐKXĐ của A và rút gọn A
b: Tìm tất cả các giá trị của x để A<-1
2: Cho 3 số dương a, b, c thỏa mãn \(a+b+c\le3\)Tìm giá trị nhỏ nhất của biểu thức
\(P=\frac{1}{a^2+b^2+c^2}+\frac{2018}{ab+bc+ca}\)
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
Cho a,b,c thỏa mãn \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1\)Tính:
H=\(\frac{\sqrt{a}-\sqrt{b}}{1+c}+\frac{\sqrt{b}-\sqrt{c}}{1+a}+\frac{\sqrt{c}-\sqrt{a}}{1+b}\)