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c)
\(\sqrt{\left(x-1\right)^2}=2\)
x-1=2
x=3
d) \(\Leftrightarrow2+3\sqrt{x}+x=x+5\)
\(\Leftrightarrow3\sqrt{x}=3\)
<=> x=1
a)
\(\Leftrightarrow\sqrt{\left(x+2\right)}.\sqrt{\left(x-2\right)}-\sqrt{x+2}=0\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+2}=0\\\sqrt{x-2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
b)
\(\Leftrightarrow\sqrt{\left(x-2\right)+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}.\sqrt{x-2}+2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
\(\Leftrightarrow2\sqrt{x-2}+\sqrt{2}-\sqrt{2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{2}\)
\(\Leftrightarrow x-2=2\)
\(\Leftrightarrow x=4\)
2 phần kia mình đăng sau (dài quá r)
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a/ ĐK: \(x \ge -1\). Đặt \(\sqrt{x+1}=a \ge 0\)
PT: \(\Leftrightarrow6a-3a-2a=5\)
\(\Leftrightarrow a=5\)
\(\Leftrightarrow x+1=15\Leftrightarrow x=24\) (nhận)
b,c: Hai ý này đều làm theo cách bình phương hoặc đưa về phương trình chứa dấu giá trị tuyệt đối được nhé.
b) Cách 1: ĐKXĐ: Tự tìm
\(\sqrt{x^{2}-4x+4}=2\Leftrightarrow x^{2}-4x+4=4\Leftrightarrow x(x-4)=0\)
\(\Leftrightarrow x=0\) hoặc \(x=4\) cả 2 cái này đều TMĐK
Cách 2: \((\sqrt{x^2-4x+4}=2)\)
\(\Leftrightarrow \sqrt{(x-2)^2}=2\)
\(\Leftrightarrow \mid x-2\mid=2\)
Với \(x\geq 2\) thì :
\(x-2=2 \Leftrightarrow x=4\) (nhận)
Với \(x<2\) thì
\(-x-2=2\Leftrightarrow x=0\) (nhận)
Vậy \(S={0;4}\)
c) Cách 1: \(\sqrt{x^{2}-6x+9}=x-2\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x^{2}-6x+9=x^{2}-4x+4 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x=\frac{5}{2} \end{matrix}\right.\)
Nghiệm TMĐK
Cách 2: \((\sqrt{x^2-6x+9}=x-2)\)
\(\Leftrightarrow \mid x-3\mid =x-2\)
Với \(x\geq 3\) thì
\(x-3=x-2\Leftrightarrow 0x=-1\) ( vô lý)
Với \(x<3\) thì
\(-x+3=x-2\Leftrightarrow -2x=-5 \Leftrightarrow x=\frac{5}{2}\)
Vậy \(S={\frac{5}{2}}\)
d) ĐKXĐ: Tự tìm
\(\sqrt{x^{2}+4}=\sqrt{2x+3}\Leftrightarrow x^{2}+4=2x+3\Leftrightarrow x^{2}-2x+1=0\Leftrightarrow (x-1)^{2}=0\)
\(\Leftrightarrow x=1\)
e) ĐKXĐ: \(x\geq \frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow \frac{2x-3}{x-1}=4\Rightarrow 2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\)
Nghiệm không TMĐK.
Phương trình vô nghiệm.
f) ĐKXĐ: \(x\geq \frac{-15}{2}\)
\(x+\sqrt{2x+15}=0\Leftrightarrow 2x+2\sqrt{2x+15}=0\Leftrightarrow 2x+15+2\sqrt{2x+15}+1-16=0\)
\(\Leftrightarrow (\sqrt{2x+15}+1)^{2}-4^{2}=0\Leftrightarrow (\sqrt{2x+15}+5)(\sqrt{2x+15}-3)=0\)
\(\Leftrightarrow \sqrt{2x+15}-3=0\Leftrightarrow \sqrt{2x+15}=3\Leftrightarrow 2x+15=9\Leftrightarrow x=-3\) (TMĐK)
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\) \(+,x\ge2\Rightarrow\left\{{}\begin{matrix}x-2\ge0\\x-1\ge1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=x-2\\\left|x-1\right|=x-1\end{matrix}\right.\Rightarrow\left|x-2\right|+\left|x-1\right|=x-2+x-1=3\Leftrightarrow2x-3=3\Leftrightarrow x=3\left(\text{t/m}\right)\) \(+,1\le x< 2\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=x-1+2-x=1\left(l\right)\) \(+,x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=-\left(x-1\right)=1-x\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=1-x+2-x=3\Leftrightarrow3-2x=3\Leftrightarrow x=0\left(\text{t/m}\right)\) \(f,\left\{{}\begin{matrix}\sqrt{x^2-9}\ge0\\\sqrt{x^2-6x+9}\ge0\end{matrix}\right.mà:\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-9}=0\\\sqrt{x^2-6x+9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\sqrt{\left(x-3\right)^2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\left|x-3\right|=0\end{matrix}\right.\Leftrightarrow x=3\)\thay vào ta thấy thoa man => x=3
\(ĐK:x\ge4\)\(\sqrt{x^2+x-20}=\sqrt{x^2+5x-4x-20}=\sqrt{x\left(x+5\right)-4\left(x+5\right)}=\sqrt{\left(x-4\right)\left(x+5\right)}=\sqrt{x-4}.\sqrt{x+5}=\sqrt{x-4}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}=0\\\sqrt{x+5}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x+5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-4\left(l\right)\end{matrix}\right.\Rightarrow x=4\) \(b,ĐK:x\le2;\sqrt{x+1}+\sqrt{2-x}=\sqrt{6}\Leftrightarrow x+1+2-x+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow3+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow2\sqrt{\left(x+1\right)\left(2-x\right)}=3\Leftrightarrow\sqrt{\left(x-1\right)\left(2-x\right)}=1,5\Leftrightarrow\left(x-1\right)\left(2-x\right)=\frac{9}{4}\Leftrightarrow\left(x-1\right)\left(x-2\right)=-\frac{9}{4}\Leftrightarrow x^2-3x+2=-\frac{9}{4}\Leftrightarrow x^2-3x+\frac{9}{4}=-2\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-2\Rightarrow vonghiem\)
1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)
<=> \(\sqrt{\left(x-10\right)^2}=10\)
<=> \(\left|x-10\right|=10\)
=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)
Vậy S = \(\left\{20;0\right\}\)
2) \(\sqrt{x +2\sqrt{x}+1}=6\)
<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)
<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)
<=> \(\left|\sqrt{x}+1\right|=6\)
=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)
Vậy S = \(\left\{25\right\}\)
3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)
<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)
<=> \(\left|x-3\right|=\sqrt{3}+1\)
=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)
Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)
4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)
<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)
<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)
<=> \(\left|\sqrt{3x}+1\right|=5\)
=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)
5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)
<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)
Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)
6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)
<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)
<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)
<=> \(\left|\sqrt{6x}+2\right|=7\)
=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)
=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)
ĐKXĐ: $x \geq 2$
\(\Leftrightarrow2\left(x-4\right).\sqrt{x-2}-2\left(x-4\right)+\left(x-2\right)\sqrt{x+1}-2\left(x-2\right)+6x-18=0\\ \Leftrightarrow2.\left(x-4\right).\dfrac{x-3}{\sqrt{x-2}+1}+\left(x-2\right).\dfrac{x-3}{\sqrt{x+1}+2}+6.\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=0\right)\\ \Leftrightarrow x=3\)
Vì \(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=\dfrac{2\left(x-4\right)+4.\sqrt{x-2}+4}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2\\ =\dfrac{2\left(x-2\right)+4.\sqrt{x-2}}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2>0\)
Vậy....