Giải phương trình: \(\frac{x^2}{\left(\sqrt{6^2-x^2}-3\right)^2}=4\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình nghĩ sửa đề bài là \(\frac{\sqrt{x^2-x+6}+7\sqrt{x}-\sqrt{6\left(x^2+5x-2\right)}}{x+3-\sqrt{2\left(x^2+10\right)}}\le0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
a:\(\sqrt{\left(\sqrt{3}-2\right)^2}\)+\(\sqrt{\left(1+\sqrt{3}\right)^2}\)
=\(\sqrt{3}-2+1+\sqrt{3}\)
=\(2\sqrt{3}-1\)
b; dài quá mink lười làm thông cảm
bài 2:
\(\sqrt{x^2-2x+1}=7\)
=>\(\sqrt{\left(x-1\right)^2}=7
\)
=>\(\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\)
=>\(\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)
b: \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
=>\(\sqrt{4\left(x-5\right)}-9\sqrt{x-5}=\sqrt{1-x}\)
\(=2\sqrt{x-5}-9\sqrt{x-5}=\sqrt{1-x}\)
=>\(-7\sqrt{x-5}=\sqrt{1-x}\)
=\(-7.\left(x-5\right)=1-x\)
=>\(-7x+35=1-x\)
=>\(-7x+x=1-35\)
=>\(-6x=-34\)
=>\(x\approx5.667\)
mink sợ câu b bài 2 sai đó bạn
1 a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(|2-\sqrt{3}|+|1+\sqrt{3}|\)
= \(2-\sqrt{3}+1+\sqrt{3}\)
= \(2+1\)= \(3\)
b) \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\cdot\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{6}{3^2}}-4\sqrt{\frac{6}{2^2}}\right)\cdot\left(3\sqrt{\frac{6}{3^2}}-\sqrt{6}\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\cdot\left(\frac{3}{3}\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}\right)\cdot\left(\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\right)\cdot\left(\sqrt{6}\left(1-\sqrt{2}-1\right)\right)\)
= \(\sqrt{6}\frac{1}{6}\cdot\sqrt{6}\left(-\sqrt{2}\right)\)
= \(\sqrt{6}^2\left(\frac{-\sqrt{2}}{6}\right)\)
= \(6\frac{-\sqrt{2}}{6}\)=\(-\sqrt{2}\)
2 a) \(\sqrt{x^2-2x+1}=7\)
<=> \(\sqrt{x^2-2x\cdot1+1^2}=7\)
<=> \(\sqrt{\left(x-1\right)^2}=7\)
<=> \(|x-1|=7\)
Nếu \(x-1>=0\)=>\(x>=1\)
=> \(|x-1|=x-1\)
\(x-1=7\)<=>\(x=8\)(thỏa)
Nếu \(x-1< 0\)=>\(x< 1\)
=> \(|x-1|=-\left(x-1\right)=1-x\)
\(1-x=7\)<=>\(-x=6\)<=> \(x=-6\)(thỏa)
Vậy x=8 hoặc x=-6
b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
<=> \(\sqrt{4\left(x-5\right)}-3\frac{\sqrt{x-5}}{3}=\sqrt{1-x}\)
<=> \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\sqrt{x-5}=\sqrt{1-x}\)
ĐK \(x-5>=0\)<=> \(x=5\)
\(1-x\)<=> \(-x=-1\)<=> \(x=1\)
Ta có \(\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\left(\sqrt{x-5}\right)^2=\left(\sqrt{1-x}\right)^2\)
<=> \(x-5=1-x\)
<=> \(x-x=1+5\)
<=> \(0x=6\)(vô nghiệm)
Vậy phương trình vô nghiệm
Kết bạn với mình nha :)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)
Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)
Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)
\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)
\(\Leftrightarrow b=a\)
Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)
\(\Leftrightarrow x^3-4x^2-6x+5=0\)
\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Vì mẫu số phải \(\ne0\) nen \(\left(\sqrt{6^2-x^2}-3\right)^2\ne0\)
\(< =>\sqrt{6^2-x^2}-3\ne0\)
\(< =>\sqrt{36-x^2}\ne3\)
\(< =>36-x^2\ne9\)
\(< =>x^2\ne27\)
\(< =>x\ne\pm3\sqrt{3}\) ( phần này bạn làm ở ngoài giấy nháp nha )
Điều kiện xác định : \(x\ne3\sqrt{3}\) \(va\) \(x\ne-3\sqrt{3}\)
\(\frac{x^2}{\left(\sqrt{6^2-x^2-3}\right)^2}=4\)
\(< =>\frac{x^2}{\left(\sqrt{6x^2-x^2}-3\right)^2}=\frac{4\left(\sqrt{6^2-x^2}-3\right)^2}{\left(\sqrt{6^2-x^2}-3\right)^2}\)
\(< =>x^2=4\left(\sqrt{6^2-x^2}-3\right)^2\)
\(< =>x^2=4.\left[\left(\sqrt{36-x^2}\right)^2-2\sqrt{36-x^2}.3+9\right]\)
\(< =>x^2=4.\left[\left(36-x^2\right)-\sqrt{6^2.\left(36-x^2\right)}+9\right]\)
\(< =>x^2=4.\left(36-x^2\right)-4.\sqrt{\left(1296-36x^2\right)}+4.9\)
\(< =>x^2=144-4x^2-\sqrt{4^2.\left(1296-36x^2\right)}+36\)
\(< =>x^2=144-4x^2-\sqrt{20736-576x^2}+36\)
\(< =>x^4=20736-16x^4-\left(20736-576x^2\right)+1296\)
\(< =>x^4=20736-16x^4-20736+576x^2+1296\)
\(< =>x^4+16x^4-576x^2-20736+20736-1296=0\)
\(< =>17x^4-576x^2-1296=0\)
\(\left(a=17;b=576;b'=288;c=-1296\right)\)
\(\Delta'=b'^2-ac\)
\(=288^2-17.\left(-1296\right)\)
\(=82944+22032\)
\(=104976\) \(>0\)
\(\sqrt{\Delta'}=\sqrt{104976}=324\)
Vậy phương trình có 2 nghiệm phân biệt
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-288+324}{17}=\frac{36}{17}\) ( nhận )
\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-288-324}{17}=-36\) ( nhận )
CHÚC BẠN HỌC TỐT !!!
x=18/5;-6