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Bài 1: Giải phương trình
a) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)
\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)
\(\Leftrightarrow100\cdot\left|x-3\right|=20\)
\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{16}{5}\right\}\)
b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
Vậy: S={10;-4}
c) Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
bài 2
ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)
\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có;
\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)
\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)
Dấu \(=\)xảy ra khi \(a=b=c=1\)
a/ \(\Delta=\left(3\sqrt{3}\right)^2-4.4\left(-6\right)=123\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{3\sqrt{3}+\sqrt{123}}{8}\\x_2=\frac{3\sqrt{3}-\sqrt{123}}{8}\end{matrix}\right.\)
b/ \(\Delta=9-4\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)=25\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{3+\sqrt{25}}{2\left(1-\sqrt{5}\right)}=-1-\sqrt{5}\\x_2=\frac{3-\sqrt{25}}{2\left(1-\sqrt{5}\right)}=\frac{1+\sqrt{5}}{4}\end{matrix}\right.\)
\(a)4x^2-3\sqrt{3}x-6=0\)
Có: \(a=4;b=-3\sqrt{3};c=-6\)
\(\Delta=b^2-4ac\\ =\left(-3\sqrt{3}\right)^2-4.4.\left(-6\right)\\ =123>0\)
Phương trình có 2 nghiệm phân biệt:
\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-3\sqrt{3}\right)+\sqrt{123}}{2.4}=\frac{3\sqrt{3}+\sqrt{123}}{8}\)
\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-3\sqrt{3}\right)-\sqrt{123}}{2.4}=\frac{3-\sqrt{123}}{8}\)
\(b)\left(1-\sqrt{5}\right)x^2-3x+\sqrt{5}+1=0\)
Có: \(a=1-\sqrt{5};b=-3;c=\sqrt{5}+1\)
\(\Delta=b^2-4ac\\ =\left(-3\right)^2-4.\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\\ =25>0\)
Phương trình có 2 nghiệm phân biệt:
\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-3\right)+\sqrt{25}}{2\left(1-\sqrt{5}\right)}=-1-\sqrt{5}\\ x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-3\right)-\sqrt{25}}{2\left(1-\sqrt{5}\right)}=\frac{1+\sqrt{5}}{4}\)