a)\(\frac{\frac{x-3}{5}+1}{4}=\frac{\frac{2x}{3}-\frac{1}{2}}{6}\)

\(\Rightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x}{6}-\frac{3}{6}}{6}\)

\(\Rightarrow\frac{\frac{x+2}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\)

\(\Rightarrow\frac{x+2}{5}:4=\frac{4x-3}{6}:6\)

\(\Rightarrow\frac{x+2}{5}.\frac{1}{4}=\frac{4x-3}{6}.\frac{1}{6}\)

\(\Rightarrow\frac{x+2}{20}=\frac{4x-3}{36}\)

\(\Rightarrow36.\left(x+2\right)=20.\left(4x-3\right)\)

\(\Rightarrow36x+72=80x-60\)

\(\Rightarrow72+60=80x-36x\)

\(\Rightarrow132=44x\)

\(\Rightarrow x=\frac{132}{44}=3\)

Vậy x=3

b)Bn làm tương tự phần a nha

Chúc bn học tốt

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Chúc bn học tốt

\(\Leftrightarrow\frac{3x-3+5}{20}=|\frac{4x-3}{36}|\)

\(\Leftrightarrow\frac{3x+2}{5}=|\frac{4x-3}{9}|\)

\(\Leftrightarrow\left(3x+2\right)9=5|4x-3|\)

ĐK: \(3x+2\ge0\Rightarrow x\ge\frac{-2}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}5.\left(4x-3\right)=9.\left(3x+2\right)\\-5.\left(4x-3\right)=9.\left(3x+2\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}20x-15=27x+18\\-20x+15=27x+18\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}7x=-33\\-47x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-33}{7}\left(L\right)\\x=\frac{-3}{47}\end{cases}}}\)

Vậy x= - 3/47

b) \(\Leftrightarrow\frac{10x+x+2}{18}=\frac{x+3+75}{60}-2\)

\(\Leftrightarrow\frac{11x+2}{18}=\frac{x-42}{60}\)

\(\Leftrightarrow10.\left(11x+2\right)=3.\left(x-42\right)\)

\(\Leftrightarrow110x+20=3x-126\)

\(\Leftrightarrow107x=-146\Leftrightarrow x=\frac{-146}{107}\)

Ta có :  \(a+b=2\)

\(\Rightarrow\)\(a = 2 -b\)

\(A = 2a^2 +3b^2 +3ab\)

\(A = 2a^2 + 3b. (a+b)\)

\(A = 2. (2-b)^2+3b. (2-b+b)\)

\(A = 2. ( b^2 -4b+4)+6b\)

\(A = 2b^2 -8b+8+6b\)

\(A = 2b^2 -2b+8\)

\(A = 2. ( b ^2 -b+4)\)

\(A=2. (b^2 -2.b.{1\over2}+({1\over2})^2-({1\over2})^2+4)\)

\(A = 2. [ (b -{1\over2})^2-{15\over4}]\)

\(A =2. (b-{1\over2})^2 + {15\over2}\)\(\ge\)\({15\over2}\)

\(Min A ={15\over2}\)\(\Leftrightarrow\)\(a = {3\over2};b={1\over2}\)

Ta có : a+b=2→b=2−a

→P=2a2+3b2+3ab=2a2+3b(a+b)=2a2+3b.2=2a2+6b=2a2+6(2−a)=2a2−6a+12

→P=2(a2−3a)+12

→P=2(a2−2a.32+94)+152

→P=2(a−32)2+152≥152

→GTNNP=152

Dấu  = xảy ra khi a−32=0

bạn ơi bạn kiểm tra lại đề thêm lần nữa xem có sai ko ?

câu a mình rút gọn ra:

\(A=\frac{5-3x}{\left(2x-3\right)\left(x-1\right)}.\frac{x}{5+3x}\)

tới đây hết rút được rồi

MỌI NGƯỜI GIÚP MIK NHA!

ĐỀ ĐÂY Ạ

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm5\end{cases}}\)

\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\)

\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x+5\right)\left(x-5\right)}\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{\left(x^2+10x+25\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)x}\)

\(\Leftrightarrow M=\frac{\left(x+5\right)^2}{x\left(x+5\right)}\)

\(\Leftrightarrow M=\frac{x+5}{x}\)

b) Để \(M\inℤ\)

\(\Leftrightarrow x+5⋮x\)

\(\Leftrightarrow5⋮x\)

\(\Leftrightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Mà \(x\ne\pm5\)

\(\Leftrightarrow x\in\left\{1;-1\right\}\)

Vậy để \(M\inℤ\Leftrightarrow x\in\left\{1;-1\right\}\)

\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\left(x\ne\pm5;x\ne0\right)\)

\(\Leftrightarrow M=\left(\frac{x}{x+5}+\frac{5}{x-5}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\left(\frac{x^2-5x}{\left(x-5\right)\left(x+5\right)}+\frac{5x+25}{\left(x-5\right)\left(x+5\right)}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)

\(\Leftrightarrow M=\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)

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

b) M là số nguyên thì x+5 chia hết cho x

=> 5 chia hết cho x

x nguyên => x thuộc Ư (5)={-5;-1;1;5}
Vậy x={-5;-1;1;5} thì M là số nguyên

x⁴ - 4x² + 12x - 9 = 0

<=> x⁴ - x³ + x³ - x² - 3x² + 3x + 9x - 9 = 0

<=> x³(x - 1) + x²(x - 1) - 3x(x - 1) + 9(x - 1) = 0

<=> (x - 1)(x³ + x² - 3x + 9) = 0

<=> (x - 1)(x³ + 3x² - 2x² - 6x + 3x + 9) = 0

<=> (x - 1)[ x²(x + 3) - 2x(x + 3) + 3(x + 3) ] = 0

<=> (x - 1)(x + 3)(x² - 2x + 3) = 0

<=> (x - 1)(x + 3)(x² - 2x + 1 + 2) = 0

<=> (x - 1)(x + 3)[ (x - 1)² + 2 ] = 0

<=> (x - 1)(x + 3) = 0 --> do (x - 1)² + 2 > 0 với mọi x

<=>

[ x - 1 = 0 =>[ x = 1

[ x + 3 = 0 =>[ x = -3

Bạn nên sửa >= là = vì giải bất phương trình mà