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\(a.x+2\sqrt{x}=\sqrt{x}\left(\sqrt{x}+2\right)\)
\(b.x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(c.x\sqrt{x}+x-\sqrt{x}-1=x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)=\left(x-1\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2\)
\(d.a-\sqrt{ab}=\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\)
\(e.x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2=\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\)
\(f.x-4\sqrt{x}+4=\left(\sqrt{x}-2\right)^2=\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)\)
\(g.x-4=\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\)
\(h.a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\)
\(i.x\sqrt{x}-27=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)\)
\(k.x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(j.x^2-\sqrt{x}=\sqrt{x}\left(x\sqrt{x}-1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}+\sqrt{x}\right)-\left(x+1\right)\)
\(=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(\sqrt{x}-1\right)\)
\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)
\(=\left(1+\sqrt{x}\right)^2-\left(2\sqrt{\sqrt{x}}\right)^2\)
\(=\left(1+\sqrt{x}+2\sqrt{\sqrt{x}}\right)\left(1+\sqrt{x}-2\sqrt{\sqrt{x}}\right)\)
\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)
\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)
\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
\(\text{f)}x-2\sqrt{x-1}-a^2\)
\(=\left(\sqrt{x-2}\right)^2\left(\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2}\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}+a\right)\left(\sqrt{x-2\sqrt{x-1}}-a\right)\)
\(a,x-3\sqrt{x}+2\)
\(=x-3\sqrt{x}+\frac{9}{4}-\frac{1}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+2\right)\left(x-2\right)\)
câu a mình nhìn nhầm :
\(=\left(x-1\right)\left(x+2\right)\)
a,\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)=\sqrt{a}\left(\sqrt{b-1}\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right).\left(\sqrt{b}-1\right).\)
mầy phần này dễ mà mình gại đánh máy quá
những phần sau sử dụng hằng đẳng thức nhé
a) Do VT >=0 nên VP >=0 nên \(x\ge4\)
\(PT\Leftrightarrow\left(x-2\right)-\sqrt{x-2}-2=0\)
Đặt \(\sqrt{x-2}=t\ge\sqrt{4-2}=\sqrt{2}\) thì \(t^2-t-2=0\)
\(\Leftrightarrow t=2\left(loại t = -1 vì nó không thỏa mãn đk\right)\Leftrightarrow x-2=4\Leftrightarrow x=6\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) ( SỬA ĐỀ)
\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)
\(|x-1-2|+|x-1-3|=1\)
\(|x-3|+|x-4|=1\)
Với \(x\le3\)thì PT thành \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)
Với \(3\le x< 4\)thì PT thành \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)
Với \(x\ge4\)thì PT thành \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)
Vậy \(3\le x\le4\)
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 \(x\ge2\)
Giải pt :
1
a. ĐKXĐ : \(x\ge4\)
Ta có :
\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)
\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)
\(\Leftrightarrow x=13\) (TM ĐKXĐ)
Vậy \(S=\left\{13\right\}\)
b.ĐKXĐ : \(-3\le x\le10\)
Ta có :
\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy \(S=\left\{1;6\right\}\)
d: \(a-\sqrt{ab}=\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\)
e: \(=\left(\sqrt{x}+1\right)^2\)
i: \(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)\)
k: \(=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)