a: \(\left\{{}\begin{matrix}3x-2y=5\\-2x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-2y=5\\-4x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x=11\\-2x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-11\\y=2x+3=-22+3=-19\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=4\\x-\dfrac{y}{2}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-y=4\\2x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=0\left(luônđúng\right)\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y\in R\\2x=y+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\in R\\x=\dfrac{1}{2}y+2\end{matrix}\right.\)

Vậy: \(\left\{{}\begin{matrix}y\in R\\x=\dfrac{y+4}{2}\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+2y=-2\\5x+4y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x+4y=-4\\5x+4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-5x=-4-1=-5\\5x+4y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\4y=1-5x=1+25=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{26}{4}=\dfrac{13}{2}\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}2x-y=6\\3x+5y=22\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10x-5y=30\\3x+5y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=52\\2x-y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=4\\2x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2x-6=2\cdot4-6=2\end{matrix}\right.\)

e: \(\left\{{}\begin{matrix}-x+2y-6=0\\5x-3y-5=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+10y=30\\5x-3y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7y=35\\x-2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=2y-6=10-6=4\end{matrix}\right.\)

g: \(\left\{{}\begin{matrix}2x-3y=8\\5x+2y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-6y=16\\15x+6y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=19\\2x-3y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\3y=2x-8=2-8=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

giúp mik với

7:

a: ĐKXĐ: x>=0; x<>1

\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

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

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

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

b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)

c: |D|=1/3

=>D=-1/3 hoặc D=1/3

=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

=>\(\sqrt{x}+1=3\)

=>\(\sqrt{x}=2\)

=>x=4

6:

a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)

\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

b: C<-1

=>C+1<0

=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

=>\(-\sqrt{x}+4< 0\)

=>\(-\sqrt{x}< -4\)

=>\(\sqrt{x}>4\)

=>x>16

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

Để `C < -1` Ta có :

 \(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)

đỉnh z

a: Thay x=2 và y=0 vào (d), ta được:

2(2-m)+m+1=0

=>4-2m+m+1=0

=>5-m=0

=>m=5

b: Đề sai rồi bạn

c: Thay x=2 vào y=-2x+3, ta được:

\(y=-2\cdot2+3=-4+3=-1\)

Thay x=2 và y=-1 vào (d), ta được:

2(2-m)+m+1=-1

=>4-2m+m+1=-1

=>-m+5=-1

=>-m=-6

=>m=6

d: Thay y=-2 vào y=2x-3, ta được:

2x-3=-2

=>2x=1

=>x=1/2

Thay x=1/2 và y=-2 vào (d), ta được:

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

=>\(-1+\dfrac{1}{2}m+m+1=-2\)

=>\(\dfrac{3}{2}m=-2\)

=>\(m=-2:\dfrac{3}{2}=-\dfrac{4}{3}\)

e: Thay x=1 và y=2 vào (d), ta được:

\(1\left(2-m\right)+m+1=2\)

=>2-m+m+1=2

=>3=2(vô lý)

phần b sai chỗ nào v bn

\(21,B\\ 22,B\\ 23,C\\ 24,A\\ 25,B\\ 16,C\\ 17,C\\ 18,D\\ 19,B\\ 20,A\)

Hóa bạn qua bên box Hóa đăng nhé

a: Áp dụng định lí Pytago vào ΔBAC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=10^2+15^2=325\)

hay \(BC=5\sqrt{13}\left(cm\right)\)

Xét ΔBAC vuông tại A có 

\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{15}{5\sqrt{13}}=\dfrac{3}{\sqrt{13}}\)

\(\Leftrightarrow\widehat{B}\simeq56^0\)

b: Xét ΔBAC có 

BI là đường phân giác ứng với cạnh AC

nên \(\dfrac{AI}{AB}=\dfrac{CI}{BC}\)

hay \(\dfrac{AI}{10}=\dfrac{CI}{5\sqrt{13}}\)

mà AI+CI=15cm

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{AI}{10}=\dfrac{CI}{5\sqrt{13}}=\dfrac{AI+CI}{10+5\sqrt{13}}=\dfrac{15}{10+5\sqrt{13}}=\dfrac{-2+\sqrt{13}}{3}\)

Do đó: \(AI=\dfrac{-20+10\sqrt{13}}{3}\left(cm\right)\)

em em cảm cảm ơn anh nhiều lắm ạ

a) đk: \(x\ge2\)

Ta có: \(\sqrt{x}+\sqrt{x-2}=2\sqrt{x-1}\) (đã sửa đề)

\(\Leftrightarrow x+2\sqrt{x\left(x-2\right)}=4\left(x-1\right)\)

\(\Leftrightarrow3x-4=2\sqrt{x^2-2x}\)

\(\Leftrightarrow9x^2-24x+16=4\left(x^2-2x\right)\)

\(\Leftrightarrow5x^2-16x+16=0\)

\(\Leftrightarrow5\left(x^2-\frac{16}{5}x+\frac{64}{25}\right)+\frac{16}{5}=0\)

\(\Leftrightarrow5\left(x-\frac{8}{5}\right)^2=-\frac{16}{5}\) vô lý

=> PT vô nghiệm

b) Đề chắc là: \(x^2+x+12=\sqrt{36}\)

\(\Leftrightarrow x^2+x+12-6=0\)

\(\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=-\frac{23}{4}\) vô lý

=> PT vô nghiệm