tae tae ơi khó quá hổng hiểu j hết trơn

mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được 

\(=\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}\)

\(=\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(=\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(=\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0 

<=> x + 59 = 0 

<=> x=-59 

\(x\ne0\)

\(\frac{2x+1}{x^2}+1+\frac{x^2}{2\left(3x^2+4x+2\right)}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{x^2+2x+1}{x^2}-\frac{2\left(x^2+2x+1\right)}{2\left(3x^2+4x+2\right)}=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(\frac{1}{x^2}-\frac{1}{3x^2+4x+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2=3x^2+4x+2\end{matrix}\right.\) \(\Rightarrow x=-1\)

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)

b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)

\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)

Suy ra:

\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)

\(\Leftrightarrow\)2x-x²-x+2+5x²-5 = 15x-15

\(\Leftrightarrow\)2x-x²-x+5x²-15x = -15+5-2

\(\Leftrightarrow\)4x²-14x = -12

\(\Leftrightarrow4x^2-14x+12=0\)

\(\Leftrightarrow4x^2-8x-6x+12=0\)

\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0

\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)

Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)

a) \(pt\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1-\frac{3x^2+16}{x^2+10}+2=0\)

\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0\)

\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)

\(\Leftrightarrow4-x^2=0\)(do \(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0,\forall x\))

\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(KL...\)

2x(8x - 1)²(4x - 1) = 9

<=> 512x⁴ - 256x³ + 40x² - 2x = 9

<=> 512x⁴ - 256x³ + 40x² - 2x - 9 = 0

<=> (2x - 1)(4x + 1)(64x⁴ - 16x + 9) = 0

vì 64x⁴ - 16x + 9 khác 0 nên:

<=> 2x - 1 = 0 hoặc 4x + 1 = 0

<=> x = 1/2 hoặc x = -1/4

Bài 1:

d)ĐKXĐ: \(x\ne8\)

Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)

\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)

MTC=24(x-8)

\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)

\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)

\(\Leftrightarrow348-29x=0\)

\(\Leftrightarrow-29x+348=0\)

\(\Leftrightarrow x=\frac{-348}{-29}=12\)

Vậy: x=12

e) ĐKXĐ: \(x\ne\pm1\)

Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)

MTC=4(x+1)(x-1)

\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)

\(\Leftrightarrow20x+4=0\)

\(\Leftrightarrow20x=-4\)

\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)

Vậy: x không có giá trị

g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)

\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)

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

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

\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)

MTC=2(x+1)

\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)

\(\Leftrightarrow2-x+1=0\)

\(\Leftrightarrow1-x=0\)

\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)

Vậy: x không có giá trị

\(x\ne\left\{-4;-3;-2;-1\right\}\)

\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)

\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)

\(\Leftrightarrow x\left(\frac{x}{x+1}+\frac{x+1}{x+2}-\frac{x+2}{x+3}-\frac{x+3}{x+4}\right)=0\)

\(\Leftrightarrow x\left(1-\frac{1}{x+1}+1-\frac{1}{x+2}+\frac{1}{x+3}-1+\frac{1}{x+4}-1\right)=0\)

\(\Leftrightarrow x\left(\frac{1}{x+3}+\frac{1}{x+4}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)

ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\\ \Leftrightarrow\frac{2x^2-5x+2}{\left(x-1\right)\left(x-2\right)}+\frac{3x^2-5x+2}{\left(x-1\right)\left(x-2\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4x^2-12x+8}{\left(x-1\right)\left(x-2\right)}\\ \Rightarrow2x^2-5x+2+3x^2-5x+2=x^2+4x+5+4x^2-12x+8\\ \Leftrightarrow2x^2+3x^2-x^2-4x^2-5x-5x-4x+12x=5-2-2\\ \Leftrightarrow-2x=1\\ \Leftrightarrow x=\frac{-1}{2}\left(tm\right)\)Vậy tập nghiệm của phương trình là: \(S=\left\{-\frac{1}{2}\right\}\)

ĐKXĐ: x∉{1;2}

Ta có: \(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\)

\(\Leftrightarrow\frac{\left(2x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\frac{\left(3x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4\left(x^2-3x+2\right)}{\left(x-1\right)\left(x-2\right)}\)

Suy ra: \(\left(2x-1\right)\left(x-2\right)+\left(3x-2\right)\left(x-1\right)=x^2+4x+5+4\left(x^2-3x+2\right)\)

\(\Leftrightarrow2x^2-4x-x+2+3x^2-3x-2x+2=x^2+4x+5+4x^2-12x+8\)

\(\Leftrightarrow5x^2-10x+4=5x^2-8x+13\)

\(\Leftrightarrow5x^2-10x+4-5x^2+8x-13=0\)

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow-2x=9\)

hay \(x=\frac{-9}{2}\)(tm)

Vậy: \(S=\left\{-\frac{9}{2}\right\}\)