Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo để được hỗ trợ tốt hơn. Viết ntn nhìn rất khó đọc

1: =>x-3+3x-9-2(3-x)=60

=>4x-12-6+2x=60

=>6x-18=60

=>6x=78

=>x=13

2: ĐKXĐ: x<>-1; x<>3

Ai làm đc câu nào thì làm giúp mình với ạ, cảm ơn trc:(((

\(1,3x-5x+5=-8\)

\(\Leftrightarrow-2x+5+8=0\)

\(\Leftrightarrow-2x=-13\)

\(\Leftrightarrow x=\frac{13}{2}\)

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a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

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

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

\(\Leftrightarrow x^2-2x-3=1\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

Không hiểu câu hỏi lắm :vvv

\(\text{CM vô nghiệm}\)
\(\text{a) }\left(x-2\right)^3=\left(x-2\right).\left(x^2+2x+4\right)-6\left(x-1\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6x^2+12x-6\)
\(\Leftrightarrow x^3-6x^2+12x-x^3+6x-12x=-8+8-6\)
\(\Leftrightarrow0x=-6\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{b) }4x^2-12x+10=0\)
\(\Leftrightarrow\left(4x^2-12x+9\right)+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2=-1\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{CM vô số nghiệm}\)
       \(\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^3-3x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x+1\right)\text{ (luôn luôn đúng)}\)
\(\text{Vậy }S\inℝ\)

a: \(\frac{2x+2}{5}+\frac{3}{10}<\frac{3x-2}{4}\)

=>\(\frac{4\left(2x+2\right)}{20}+\frac{6}{20}<\frac{5\left(3x-2\right)}{20}\)

=>4(2x+2)+6<5(3x-2)

=>8x+8+6<15x-10

=>8x+14<15x-10

=>-7x<-24

=>\(x>\frac{24}{7}\)

b: \(\frac{2+x}{3}<\frac{3+2x}{5}\)

=>\(\frac{x+2}{3}-\frac{2x+3}{5}<0\)

=>\(\frac{5\left(x+2\right)-3\left(2x+3\right)}{15}<0\)

=>5(x+2)-3(2x+3)<0

=>5x+10-6x-9<0

=>-x+1<0

=>-x<-1

=>x>1

d: \(1+\frac{3\left(x+1\right)}{10}>\frac{x-2}{5}\)

=>\(\frac{10+3\left(x+1\right)}{10}>\frac{2\left(x-2\right)}{10}\)

=>3(x+1)+10>2(x-2)

=>3x+3+10>2x-4

=>3x+13>2x-4

=>3x-2x>-4-13

=>x>-17

e: \(\frac{2x-7}{6}\ge\frac{3x-7}{2}\)

=>\(\frac{2x-7}{6}\ge\frac{3\left(3x-7\right)}{6}\)

=>2x-7>=3(3x-7)

=>2x-7>=9x-21

=>-7x>=-14

=>x<=2

f: \(\frac{2x-1}{3}>\frac{3x+1}{2}\)

=>\(\frac{2x-1}{3}-\frac{3x+1}{2}>0\)

=>\(\frac{2\left(2x-1\right)-3\left(3x+1\right)}{6}>0\)

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

=>4x-2-9x-3>0

=>-5x-5>0

=>5x+5<0

=>5x<-5

=>x<-1