Câu a chắc em ghi đề nhầm, H,B,E thẳng hàng nên làm sao nó là tam giác được.

HFB và HEC đồng dạng thì đúng

a.

\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)

\(=\left(\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\left(\dfrac{x^2-2x+1-\left(x^2+x\right)+3x+1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)

\(=\dfrac{2}{2x+1}\)

b.

\(Q=5P=\dfrac{10}{2x+1}\)

Do Q nguyên tố nên Q cũng là số tự nhiên \(\Rightarrow\dfrac{10}{2x+1}\) là số tự nhiên

\(\Rightarrow2x+1=Ư\left(10\right)\)

Mà 2x+1 luôn lẻ nên ta chỉ cần xét các ước tự nhiên lẻ của 10

\(\Rightarrow2x+1=\left\{1;5\right\}\)

\(\Rightarrow x=\left\{0;2\right\}\)

Với \(x=0\Rightarrow Q=10\) ko phải SNT (loại)

Với \(x=2\Rightarrow Q=2\) là SNT (thỏa mãn)

Vậy \(x=2\) thì Q là SNT

a.

Do BE, CF là các đường cao \(\Rightarrow\widehat{BFH}=\widehat{CEH}=90^0\)

Xét hai tam giác BHF và CHE có:

\(\left\{{}\begin{matrix}\widehat{BFH}=\widehat{CEH}=90^0\\\widehat{BHF}=\widehat{CHE}\left(\text{đối đỉnh}\right)\end{matrix}\right.\) 

\(\Rightarrow\Delta BHF\sim\Delta CHE\left(g.g\right)\)

\(\Rightarrow\dfrac{HB}{HC}=\dfrac{HF}{HE}\Rightarrow HE.HB=HC.HF\)

b.

Xét hai tam giác BAE và CAF có:

\(\left\{{}\begin{matrix}\widehat{A}-chung\\\widehat{BEA}=\widehat{CFA}=90^0\end{matrix}\right.\)

\(\Rightarrow\Delta BAE\sim\Delta CAF\left(g.g\right)\)

\(\Rightarrow\dfrac{AE}{AF}=\dfrac{AB}{AC}\Rightarrow\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

Xét hai tam giác AEF và ABC có:

\(\left\{{}\begin{matrix}\widehat{A}-chung\\\dfrac{AE}{AB}=\dfrac{AF}{AC}\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta AEF\sim\Delta ABC\left(c.g.c\right)\)

a) Xét \(\Delta BHF\) và \(\Delta CHE\) có:

\(\left\{{}\begin{matrix}\widehat{BFH}=\widehat{CEH}=90^{\circ}\left(CF\bot AB;BE\bot AC;BE\cap CF=\left\{H\right\}\right)\\\widehat{BHF}=\widehat{CHE}\left(\text{hai góc đối đỉnh}\right)\end{matrix}\right.\)

\(\Rightarrow \Delta BHF\backsim \Delta CHE\) \(\left(g.g\right)\Rightarrow\dfrac{HB}{HC}=\dfrac{HF}{HE}\) (các cạnh tương ứng)

\(\Rightarrow HE\cdot HB=HC\cdot HF\) (đpcm)

b) Xét \(\Delta ABE\) và \(\Delta ACF\) có: \(\left\{{}\begin{matrix}\widehat{AEB}=\widehat{AFC}=90^{\circ}\left(BE\bot AC;CF\bot AB\right)\\\widehat{BAC}\text{ chung}\end{matrix}\right.\)

\(\Rightarrow \Delta ABE \backsim \Delta ACF\) \(\left(g.g\right)\Rightarrow\dfrac{AB}{AC}=\dfrac{AE}{AF}\) (các cạnh tương ứng)

\(\Rightarrow\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

Xét \(\Delta AEF\) và \(\Delta ABC\) có: \(\left\{{}\begin{matrix}\dfrac{AE}{AB}=\dfrac{AF}{AC}\left(cmt\right)\\\widehat{BAC}\text{ chung}\end{matrix}\right.\)

\(\Rightarrow \Delta AEF \backsim \Delta ABC\) \(\left(c.g.c\right)\) (đpcm)

\(\text{#}Toru\)

Bài 1:

a, \(\left(5x^3-4x\right):\left(-2x\right)\)

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

\(=\dfrac{-5}{2}x^2+2\)

b, \(\left(-2x^5-4x^3+3x^2\right):2x^2\)

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

\(=-x^3-2x+\dfrac{3}{2}\)

c, \(\left(-5x^3+15x^2+18x\right):\left(-5x\right)\)

\(=-5x^3:\left(-5x\right)+15x^2:\left(-5x\right)+18x:\left(-5x\right)\)

\(=-x^2-3x-\dfrac{18}{5}\)

d, \(\left(-15x^6-24x^3\right):\left(-3x^2\right)\)

\(=-15x^6:\left(-3x^2\right)+\left(-24x^3\right):\left(-3x^2\right)\)

\(=5x^4+8x\)

Bài 2:

a, \(\left(x^2-2x+1\right):\left(x-1\right)\)

\(=\left(x^2-x-x+1\right):\left(x-1\right)\)

\(=\left[x\left(x-1\right)-\left(x-1\right)\right]:\left(x-1\right)\)

\(=\left(x-1\right)^2:\left(x-1\right)=x-1\)

b, \(\left(6x^3-2x^2-9x+3\right):\left(3x-1\right)\)

\(=\left[2x^2\left(3x-1\right)-3\left(3x-1\right)\right]:\left(3x-1\right)\)

\(=\left(3x-1\right)\left(2x^2-3\right):\left(3x-1\right)=2x^2-3\)

c, \(\left(x^3-4x^2-x+12\right):\left(x-3\right)\) (sửa đề)

\(=\left(x^3-3x^2-x^2+3x-4x+12\right):\left(x-3\right)\)

\(=\left[x^2\left(x-3\right)-x\left(x-3\right)-4\left(x-3\right)\right]:\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x-4\right):\left(x-3\right)=x^2-x-4\)

d, \(\left(4x^4+14x^3+21x-9\right):\left(2x^2-3\right)\)

\(=\left(4x^4-6x^2+14x^3-21x+6x^2-9+42x\right):\left(2x^2-3\right)\)

\(=\left[2x^2\left(2x^2-3\right)+7x\left(2x^2-3\right)+3\left(2x^2-3\right)+42x\right]:\left(2x^2-3\right)\)

\(=\left[\left(2x^2-3\right)\left(2x^2+7x+3\right)+42x\right]:\left(2x^2-3\right)\)

\(=2x^2+7x+3+42:\left(2x^2-3\right)\)

$\text{#}Toru$

Bài 1:

a.

$(x^4-2x^3+2x-1):(x-1)=[x^3(x-1)-x^2(x-1)-x(x-1)+(x-1)]:(x-1)$

$=[(x-1)(x^3-x^2-x+1)]:(x-1)=x^3-x^2-x+1$
b.

$-3x^3+2x^2-5x+1

$=-3x(x^2-x+1)-(x^2-x+1)-3x+2$

$=(-3x-1)(x^2-x+1)+(2-3x)

$\Rightarrow (-3x^3+2x^2-5x+1):(x^2-x+1)=-3x-1$ dư $2-3x$

c.

$x^5+4x^3+3x^2-5x+5=x^2(x^3-x+3)+5(x^3-x+3)-10$

$=(x^3-x+3)(x^2+5)-10$

$\Rightarrow (x^5+4x^3+3x^2-5x+5): (x^3-x+3)=x^2+5$ dư $-10$

\(\dfrac{x+1}{99}+\dfrac{x+4}{96}+\dfrac{x+7}{93}+\dfrac{x+9}{91}+4=0\) (sửa đề)

\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{96}+\dfrac{x+100}{93}+\dfrac{x+100}{91}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{96}+\dfrac{1}{93}+\dfrac{1}{91}\right)=0\)

\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{96}+\dfrac{1}{93}+\dfrac{1}{91}>0\))

\(\Leftrightarrow x=-100\)