giúp mk nhé

1.(-9)+301+(-511)+119

=(119-9)-(511-301)

=110-210

=-100

2.  455+(-311)+2015+(-144)

=2015+455-(311+144)

=2015 +455 - 455

=2015

\(\frac{x+2}{13}+\frac{2x+45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)

\(\Leftrightarrow\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)

\(\Leftrightarrow\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)

\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{7}-\frac{4}{9}\right)=0\)

mà \(\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{7}-\frac{4}{9}\right)\ne0\)

\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)

Vậy x=-15

k cho mình nha

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left[\left(x^2-7\right)+3\right]\left[\left(x^2-7\right)-3\right]=72\)

\(\Leftrightarrow\left(x^2-7\right)^2-9=72\)

\(\Leftrightarrow\left(x^2-7\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-7=9\\x^2-7=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=16\\x^2=-2\left(loai\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

Đặt \(y=x^2-7\) Pt trở thành: \(\left(y+3\right)\left(y-3\right)=73\)

\(\Leftrightarrow y^2-9=72\)

\(\Leftrightarrow y^2=81\)

\(\Leftrightarrow y=\pm9\)

Từ đó tìm được: \(x_1=-4;x_2=4\)

\(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)-\left(4x^2-8x-4\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm2\end{cases}}}\)

\(\left|x\right|-3x=15\)

\(\left|x\right|=15+3x\)

TH1 : x < 0 thì : 

\(x=-15-3x\)

\(x+3x=-15\)

\(4x=-15\)

\(x=-\frac{15}{4}\) ( Thỏa mãn )

Th2 : x > 0 thì : 

\(x=15+3x\)

\(x-3x=15\)

\(-2x=15\)

\(x=-\frac{15}{2}\) ( Loại )

Vậy \(x=-\frac{15}{4}\)

                                     Bài giải

\(\left|x\right|-3x=15\)

\(\left|x\right|=15+3x\)

TH1 : x < 0 thì : 

\(x=-15-3x\)

\(x+3x=-15\)

\(4x=-15\)

\(x=-\frac{15}{4}\) ( Thỏa mãn )

TH2 : x > 0 thì : 

\(x=15+3x\)

\(x-3x=15\)

\(-2x=15\)

\(x=-\frac{15}{2}\) ( Loại )

Vậy \(x=-\frac{15}{4}\)

Kẻ 1 đường phụ nối M và C. Ta có diện tích tam giác AMC là:7,5+7,5x1/4=9,375(cm2)

Vậy S tam giác ABC là: 9,375+9,375x 2/3=15,625(cm2)

kết bạn vs mình nếu câu trả lời đó đúng.

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

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

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