\(a,PT\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)\left(x+1\right)=0\Leftrightarrow\left(x-5\right)\left(x^2+x+1\right)\Leftrightarrow x=5\) (Vì \(x^2+x+1>0\forall x\))

Vậy: \(S=\left\{5\right\}\)

\(b,PT\Leftrightarrow x^6=1\Leftrightarrow x=\pm1\)

Vậy: \(S=\left\{\pm1\right\}\)

Định lí hàm cos:

\(a^2=b^2+c^2-2bc.cosA=7^2+5^2-2.7.5.\dfrac{3}{5}\Leftrightarrow a=4\sqrt{2}\)

\(\Rightarrow S_{\Delta ABC}=\dfrac{1}{2}.b.c.sinA=\dfrac{1}{2}.7.5.\dfrac{4}{5}=14\) (vì \(cosA=\dfrac{3}{5}\Rightarrow sinA=\dfrac{4}{5}\))

\(\Rightarrow h_a=\dfrac{2S_{\Delta ABC}}{a}=\dfrac{28}{4\sqrt{2}}=\dfrac{7\sqrt{2}}{2}\)

Vậy chọn đáp án \(A\).

Đặt chỗ trống cần tìm là \(x\)

Khi đó: \(\dfrac{-19}{42}+\dfrac{9}{x}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{9}{x}=\dfrac{9}{7}\)

\(\Rightarrow x=7\)

Vậy chỗ trống cần tìm là \(7\).

Gọi ô trống là \(x\)

\(\Rightarrow-\dfrac{3}{7}-\dfrac{39}{x}=-\dfrac{9}{8}\)

\(\Rightarrow\dfrac{-39x}{56}=-\dfrac{9}{8}+\dfrac{3}{7}=-\dfrac{39}{56}\)

\(\Rightarrow x=56\)

Vậy chỗ trống cần tìm là \(56\).

\(a,PT\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)

Vậy: \(S=\left\{0;0,5\right\}\)

\(b,\) Sửa đề: \(x^4+2x^3+x^2=0\)

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

Vậy: \(S=\left\{0;-1\right\}\)

\(c,PT\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\left(x^2+x+1>0\right)\)

\(\Rightarrow x=1\)

Vậy: \(S=\left\{1\right\}\)

\(d,PT\Leftrightarrow2x\left(3x-2\right)-\left(3x-2\right)=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};\dfrac{1}{2}\right\}\)

Sửa đề: CM \(\Delta ABM=\Delta ACM\)

\(a,\) Xét \(\Delta ABM\) và \(\Delta ACM\) có:

\(AM\) là cạnh chung

\(\widehat{BAM}=\widehat{CAM}\) (giả thiết)

\(BM=MC\) (giả thiết)

\(\Rightarrow\Delta ABM=\Delta ACM\left(c.g.c\right)\)

b) \(\Delta ABC\) có: \(AM\) là đường phân giác, đồng thời là đường trung tuyến (giả thiết)

\(\Rightarrow\Delta ABC\) cân tại \(A\)

\(PT\Leftrightarrow2x^3+2x^2-5x^2-5x+8x+8=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+8\right)=0\)

Chứng minh được: \(2x^2-5x+8>0\forall x\)

\(\Rightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy: \(S=\left\{-1\right\}\)

\(PT\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1-\dfrac{x+3}{7}+1-\dfrac{x+4}{6}+1=0\)

\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}=0\right)\)

Mà \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}< 0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

Vậy: \(S=\left\{-10\right\}\)

\(a,PT\Leftrightarrow\dfrac{3}{6}+\dfrac{2x-2}{6}=\dfrac{2x+1}{6}\)

\(\Leftrightarrow3+2x-2-2x-1=0\)

\(\Leftrightarrow0x=0\) (luôn đúng \(\forall x\))

Vậy phương trình có vô số nghiệm.

\(b,PT\Leftrightarrow\dfrac{7x-7}{35}+\dfrac{35x}{35}=\dfrac{5x+5}{35}\)

\(\Leftrightarrow7x-7+35x-5x-5=0\)

\(\Leftrightarrow37x=12\)

\(\Leftrightarrow x=\dfrac{12}{37}\)

Vậy: \(S=\left\{\dfrac{12}{37}\right\}\)

\(a,PT\Leftrightarrow\dfrac{3\left(x+1\right)}{6}-\dfrac{2x}{6}=\dfrac{3}{6}\)

\(\Leftrightarrow3x+3-2x=3\)

\(\Leftrightarrow x=0\)

Vậy: \(S=\left\{0\right\}\)

\(b,PT\Leftrightarrow\left(x+5\right)\left(-6-x\right)\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Vậy: \(S=\left\{-5;-6\right\}\)

\(c,\) ĐKXĐ: \(x\ne\pm1\)

\(PT\Leftrightarrow\left(x-2\right)\left(x-1\right)-x\left(x+1\right)=6-x\)

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

\(\Leftrightarrow x=-\dfrac{4}{3}\)

Vậy: \(S=\left\{-\dfrac{4}{3}\right\}\)