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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
Lời giải:
a)
ĐK: \(\forall x\in\mathbb{R}\)
Ta có: \(\sqrt{3x^2}-\sqrt{12}=0\)
\(\Rightarrow \sqrt{3x^2}=\sqrt{12}\)
\(\Rightarrow 3x^2=12\Rightarrow x^2=4\Rightarrow x=\pm 2\) (đều thỏa mãn)
b) ĐK: \(\forall x\in\mathbb{R}\)
\(\sqrt{(x-3)^2}=9\)
\(\Leftrightarrow |x-3|=9\Rightarrow \left[\begin{matrix} x-3=9\\ x-3=-9\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=12\\ x=-6\end{matrix}\right.\)
c) ĐK: $x\in\mathbb{R}$
\(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow \sqrt{(2x)^2+2.2x+1}=6\)
\(\Leftrightarrow \sqrt{(2x+1)^2}=6\)
\(\Leftrightarrow |2x+1|=6\)
\(\Rightarrow \left[\begin{matrix} 2x+1=6\\ 2x+1=-6\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=-\frac{7}{2}\end{matrix}\right.\)
d) ĐK: \(x\geq 1\)
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow \sqrt{16(x-1)}-\sqrt{9(x-1)}+\sqrt{4(x-1)}+\sqrt{x-1}=8\)
\(\Leftrightarrow 4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow 4\sqrt{x-1}=8\Rightarrow \sqrt{x-1}=2\)
\(\Rightarrow x=2^2+1=5\) (thỏa mãn)
e)
ĐK: \(-4\leq x\leq \frac{1}{2}\)
\(\sqrt{1-x}+\sqrt{1-2x}=\sqrt{x+4}\)
\(\Leftrightarrow \sqrt{1-x}-1+\sqrt{1-2x}-1=\sqrt{x+4}-2\)
\(\Leftrightarrow \frac{(1-x)-1}{\sqrt{1-x}+1}+\frac{(1-2x)-1}{\sqrt{1-2x}+1}=\frac{(x+4)-2^2}{\sqrt{x+4}+2}\)
\(\Leftrightarrow \frac{-x}{\sqrt{1-x}+1}+\frac{-2x}{\sqrt{1-2x}+1}=\frac{x}{\sqrt{x+4}+2}\)
\(\Leftrightarrow x\left(\frac{1}{\sqrt{x+4}+2}+\frac{1}{\sqrt{1-x}+1}+\frac{2}{\sqrt{1-2x}+1}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn lớn hơn $0$
Do đó: \(x=0\) là nghiệm duy nhất của pt.
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: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
\(f,\sqrt{x^2-25}-\sqrt{x-5}=0\)
=> \(\sqrt{x^2-25}=\sqrt{x-5}\)
=>\(x^2-25=x-5\)
=>\(x^2-x=25-5=20\)
=>( đến đoạn này mình xin chịu )
\(a,\sqrt{16x}=8\)
=>\(16x=8^2\)
=>\(16x=64\)
=>\(x=64:16=4\)
Vậy \(x\in\left\{4\right\}\)
\(b,\sqrt{x^2}=2x-1\)
=>\(x=2x-1\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy \(x\in\left\{1\right\}\)
\(c,\sqrt{9.\left(x-1\right)}=21\)
=>\(9.\left(x-1\right)=21^2=441\)
=> \(x-1=441:9=49\)
=>\(x=49+1=50\)
Vậy \(x\in\left\{50\right\}\)
\(d,\sqrt{4\left(1-x\right)^2}-6=0\)
=>\(\sqrt{4\left(1-x\right)^2}=0+6=6\)
=> \(4\left(1-x\right)^2=6^2=36\)
=>\(\left(1-x\right)^2=36:4=9\)
=>\(1-x=\sqrt{9}=3\)
=>\(x=1-3=-2\)
Vậy \(x\in\left\{-2\right\}\)
\(g,\sqrt{9\left(2-3x\right)^2}=6\)
=> \(9.\left(2-3x\right)^2=6^2=36\)
=> \(\left(2-3x\right)^2=36:9=4\)
=> \(2-3x=\sqrt{4}=2\)
=>\(3x=2-2=0\)
=>\(x=0:3=0\)
Vậy \(x\in\left\{0\right\}\)
( còn các bài còn lại mình sẽ nghĩ tiếp , HS6-7 làm bài )
Bài 4 :
\(a,\sqrt{x-1}=2\)
=> \(x-1=2^2=4\)
=>\(x=4+1=5\)
Vậy \(x\in\left\{5\right\}\)
\(b,\sqrt{x^2-3x+2}=2\)
=> \(x^2-3x+2=2\)
=> \(x^2-3x=2-2=0\)
=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )
=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
MÌNH Biết vậy thôi ,
Bài 4 :
c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)
\(\Leftrightarrow4x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-4x-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
\(\Leftrightarrow2=2\)( luôn đúng )
+) Xét \(1\le x< 2\):
\(pt\Leftrightarrow\sqrt{x-1}+1-1+\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)( loại )
Vậy \(x\ge2\)
\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{\left(\sqrt{8}-\sqrt{3}\right)\left(\sqrt{8}+\sqrt{3}\right)}\)
\(=\sqrt{3}+1+\sqrt{3}-1+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{5}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{5}\)
\(=2\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{8}+\sqrt{3}\)
\(=4\sqrt{3}\)
Giải pt:
1/ \(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Rightarrow x=3\)
2/ \(\Leftrightarrow\sqrt{3}x^2=\sqrt{12}\Leftrightarrow x^2=\sqrt{4}=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
3/ \(\Leftrightarrow x-5=9\Rightarrow x=14\)
4/ Đề thiếu
5/ \(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
6/ \(\Leftrightarrow2\left|1-x\right|=6\)
\(\Leftrightarrow\left|1-x\right|=3\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
7/ \(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow x-1=49\Rightarrow x=50\)
8/ \(\Leftrightarrow x+1=2^3=8\)
\(\Rightarrow x=7\)
9/ \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
10/ \(\Leftrightarrow\sqrt{2}x=\sqrt{50}\Leftrightarrow x=\sqrt{25}\Rightarrow x=5\)
11/ \(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
12/ \(\Leftrightarrow3-2x=\left(-2\right)^3=-8\)
\(\Leftrightarrow2x=11\Rightarrow x=\frac{11}{2}\)
\(\sqrt{4x^2}=6\Rightarrow\left|2x\right|=6\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-6\end{matrix}\right.\) \(\Rightarrow x=\pm3\)
b/ ĐKXĐ: \(x\ge0\)
\(\sqrt{16x}=8\Leftrightarrow16x=64\Rightarrow x=4\)
c/ ĐKXĐ: \(x\ge1\)
\(\sqrt{9\left(x-1\right)}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x-1=49\Rightarrow x=50\)
d/ \(\sqrt{4\left(1-x\right)^2}=6\Leftrightarrow2\left|1-x\right|=6\Leftrightarrow\left|1-x\right|=3\Rightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
e/ \(\sqrt{1-4x+4x^2}=5\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
f/ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\sqrt{9x^2}=2x+1\Leftrightarrow\left|3x\right|=2x+1\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\\-3x=2x+1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{5}\end{matrix}\right.\)