Đề sai rồi bạn

\(k,=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)+5\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}+5}\\ =\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}+5\right)}{\sqrt{a}+\sqrt{b}+5}=\sqrt{a}-\sqrt{b}\)

\(h,=\dfrac{1}{2a-1}\sqrt{25a^2\left(a^2-4a+4\right)}=\dfrac{1}{2a-1}\sqrt{25a^2\left(a-2\right)^2}\\ =\dfrac{\left|5a\left(a-2\right)\right|}{2a-1}=\left[{}\begin{matrix}\dfrac{5a\left(a-2\right)}{2a-1}\left(a\ge2;a\ne\dfrac{1}{2}\right)\\\dfrac{5a\left(2-a\right)}{2a-1}\left(0\le a< 2;a\ne\dfrac{1}{2}\right)\\\dfrac{-5a\left(2-a\right)}{2a-1}\left(a< 0\right)\end{matrix}\right.\)

e: vecto AM=(x-3;y+1)

vecto BM=(x+1;y-2)

vecto AC=(-2;0)

vecto AM=2*vecto BM-3*vecto AC

=>x-3=2*(x+1)+6 và y+1=2(y-2)

=>x-3=2x+8 và y+1=2y-4

=>x=-11 và y=5

f: Tọa độ H là:

\(\left\{{}\begin{matrix}x=\dfrac{3-1+1}{3}=1\\y=\dfrac{-1+2-1}{3}=0\end{matrix}\right.\)

g: K thuộc Oy nên K(0;y)

vecto AB=(-4;3)

vecto AK=(-3;y+1)

A,K,B thẳng hàng

=>\(-\dfrac{3}{-4}=\dfrac{y+1}{3}\)

=>y+1=9/4

=>y=5/4

h: P thuộc Ox nên P(x;0)

vecto PA=(x-3;1)

vecto PC=(x-1;1)

ΔPAC vuông tại P

=>vecto PA*vecto PC=0

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

=>x^2-4x+3+1=0

=>x=2

=>P(2;0)

Đề yêu cầu gì em, thu gọn đa thức hả?

\(e,=6x^2-3x+2x-1+9x+12-6x^2-8x\\ =\left(6x^2-6x^2\right)+\left(-3x+2x+9x-8x\right)+\left(-1+12\right)\\ =11\\ g,=\left(3x-3\right)\left(x-2\right)-\left(3x^2+x\right)\left(1-x\right)\\ =3x^2-3x-6x+6-\left(3x^2+x-3x^3-x^2\right)\\ =3x^2-9x+6+3x^3-2x^2-x\\ =3x^3+x^2-10x+6\)

\(e,\left(x-2\right)^2-16=0\\ \Leftrightarrow\left(x-6\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ f,x^2-5x-14=0\\ \Leftrightarrow\left(x-7\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ g,8x\left(x-3\right)+x-3=0\\ \Leftrightarrow\left(8x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=3\end{matrix}\right.\)

e)\(\left(x-2-4\right)\left(x-2+4\right)=\left(x-6\right)\left(x+2\right)\)

f)\(x^2-5x-14=x^2-2.\dfrac{5}{2}x+\dfrac{25}{2}+\dfrac{3}{2}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{2}\)

e: Ta có: \(\left(x-2\right)^2-16=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

f: Ta có: \(x^2-5x-14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)

Bài 1:

Theo đề ta có : A₁=G₁; T₁=3A₁; X₁ = 2T₁

=>\(\left\{{}\begin{matrix}X_1=6A_1\\T_1=3A_1\\G_1=A_1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}A=4A_1\\G=7A_1\end{matrix}\right.\)

Lại có: 2A + 3G= 5800

=> 29A₁ = 5800 => A₁=200

=> G = 1400