a/ ĐKXĐ: \(x\ge1\)

Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm

b/ \(x\ge1\)

\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)

Đặt \(\sqrt{x-1}=a\ge0\) ta được:

\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)

- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)

- Với \(0\le a\le1\) ta được:

\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)

- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)

c/ ĐKXĐ: \(x\ge\frac{49}{14}\)

\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)

Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(7-\sqrt{14x-49}\ge0\)

\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)

Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)

d/ ĐKXĐ: \(x\ge\frac{1}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}-2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|-2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=4\)

TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=4\)

\(\Leftrightarrow\sqrt{2x-1}=5\)

\(\Leftrightarrow x=13\)

TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)

TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)

\(\sqrt{2x-1}-1-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow4=4\) (luôn đúng)

TH4: \(\frac{1}{2}\le x< 1\)

\(1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(l\right)\)

Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ge-1\)

\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)

\(\Leftrightarrow...\)

a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+2a^2=-b^2+b+3ab\)

\(\Leftrightarrow\left(2a^2-3ab+b^2\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a+1=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x+5+4\sqrt{x+1}=1-x\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)

\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)

\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

\(\Leftrightarrow x\sqrt{2x+3}-\sqrt{2x+3}+3-3x+3\sqrt{x+5}-\sqrt{\left(2x+3\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\sqrt{2x+3}\left(x-1\right)-3\left(x-1\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\sqrt{2x+3}-3\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left(x-1-\sqrt{x+5}\right)\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{x+5}=0\\\sqrt{2x+3}-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5-\sqrt{x+5}-6=0\\\sqrt{2x+3}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-2\left(loại\right)\\\sqrt{x+5}=3\\\sqrt{2x+3}=3\end{matrix}\right.\)

\(\Leftrightarrow...\)

1) 

a) \(\sqrt{2x-4}\) có nghĩa khi:

\(2x-4\ge0\)

\(\Leftrightarrow2x\ge4\)

\(\Leftrightarrow x\ge\dfrac{4}{2}\)

\(\Leftrightarrow x\ge2\)

b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi 

\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)

\(\Rightarrow4-x\le0\)

\(\Leftrightarrow x\ge4\)

2) 

a) \(A=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(A=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\cdot\sqrt{5}+2^2}\)

\(A=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(A=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)

\(A=\sqrt{5}+2-\sqrt{5}+2\)

\(A=4\)

\(B=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-5}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{5}-\sqrt{7}}\)

\(B=\left(-\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(B=\left[-\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=\left(-\sqrt{7}-\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

\(B=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=-\left(7-5\right)\)

\(B=-2\)

1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

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)

giải bài nào hộ mk cx được ko cần lm hết đâu :) :) :)

a.

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

c.

ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)

\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)

\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=-1\)

d.

ĐKXĐ: \(x>1\)

\(\Leftrightarrow\dfrac{x^2+x+1-1}{\sqrt{x^2+x+1}}=\dfrac{1-\left(x-1\right)}{\sqrt{x-1}}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2-1}{a}=\dfrac{1-b^2}{b}\)

\(\Leftrightarrow a-\dfrac{1}{a}=\dfrac{1}{b}-b\)

\(\Leftrightarrow a+b-\dfrac{a+b}{ab}=0\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow1-\dfrac{1}{ab}=0\)

\(\Leftrightarrow ab=1\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=1\)

\(\Leftrightarrow x^3-1=1\)

\(\Leftrightarrow x=\sqrt[3]{2}\)