ta có:\(\frac{a^2-4}{2x-5}=2+a\)
\(ĐKXĐ:x\ne\frac{5}{2}\)
\(\Rightarrow\left(2+a\right).\left(2x-5\right)=a^2-4\)
\(\Rightarrow2x-5=\frac{a^2-4}{a+2}=a-2\)
\(\Leftrightarrow x=\frac{a-3}{2}\)
vì x là số nguyên dương nhỏ hơn 2 nên x=1
\(\Leftrightarrow1=\frac{a-3}{2}\)
\(\Leftrightarrow a-3=2\)
\(\Leftrightarrow a=5\)
cho phan thuc 4x-4/2x^2-2 :
a) Tim dieu kiem xac dinh
b)cho gia tri cua x=-2 tinh phan thuc
c) Tim gia tri cua x de gia tri phan thuc la so nguyen
gia tri cua a de pt a(x-1)/5+6x-17/10+2-3x/4=0 vo nghiem la a=?
cho P= 1 : a^2 +a+1
Tim nhung gia tri cua a de 1:P nho nhat va tim nhung gia tri do
1) Cho bieu thuc A=\(3+\frac{2}{x-1}\). Tinh gia tri cua bieu thuc A khi |2x-3|=1
2) Rut gon bieu thuc B=\(\frac{x}{x-1}\)-\(\frac{x-5}{x+1}\)-\(\frac{3-x}{1-x^2}\)
3) Tim cac gia tri nguyen cua x de bieu thuc \(\frac{B}{A}\)co gia tri nguyen duong
1. cho phan tuc A=\(\frac{x^2+5x+4}{x^2+x-12}\)tap hop cac gia tri cua x de phan thuc A ko xac dinh la
2.cho hinh thang ABCD co dien tich la18cm2. duong cao AH =3cm . Do dai duong trung binh hinh thang la
3.cho x, y nguyen thoa man x2+4x+y2-12=0 . khi do gia tri lon nhat P=x2+y2 la Pmax=?
4.gia tri nho nhat cua bieu thuc A= \(\frac{x^2}{4}+3x+y^2-\frac{14}{5}y+\frac{425}{25}la\)
a) cho vi du ve 1 don thuc ,1 da thuc nhung k phai la don thuc , 2 don thuc dong dang voi nhau .da thuc 1 bien
b) Hay viet 1 don thuc dong dang voi don thuc xy\(^3\) sao cho tai x=-1 y=2 gia tri cua don thuc do la so nguyen duong nho hon 10
GIUP MINH NHA
cho bieu thuc : M ( x phan x^2 -25 - x-5 phan x^2 +5x ) : 2x-5 phan 5x - x^2
a) rut gon M
b) tinh gia tri cua M voi x = -2010
c) tim tat ca cac gia tri nguyen cua x de M co gia tri nguyen.
a) Cho vidu ve 1 don thuc ,moi da thuc nhung k phai la don thuc 2 don thuc dong dang voi nhau,da thuc 1 bien
b) Hay viet 1 don thuc dong dang voi don thuc xy\(^3\) sao cho tai x=-1 y=2 gia tri cua don thuc do la so nguyen duong nho hon 10
CAC BAN GIUP MINH NHA MINH GAP LAM
cho hai so duong xy thoa man \(\frac{4}{x^2}+\frac{5}{y^2}\ge9\) tim gia tri nho nhat cua bieu thuc\(Q=2x^2+\frac{6}{x^2}+3y^2+\frac{8}{y^2}\)
