BÀi 1 :
BÀi 1 :
\(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\div\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left[\dfrac{\left(\sqrt{a}\right)-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\div\left[\dfrac{\left(a-1\right)-\left(a-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right]\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\times\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{1}\)
\(=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
(^~^)
\(\dfrac{\sqrt{a}-2}{\sqrt{a}}>\dfrac{1}{6}\)
\(\Leftrightarrow6\sqrt{a}-12>\sqrt{a}\)
\(\Leftrightarrow5\sqrt{a}>12\)
\(\Leftrightarrow a>\dfrac{144}{25}\)
Bài 1:
\(P=1+\left(\dfrac{2a+\sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\right)\)
\(=1+\left[\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(2\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+a+1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(2\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)\(\times\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{a\left(\sqrt{a}+1\right)}{\sqrt{a}+a+1}-\sqrt{a}\right)\)
\(=\dfrac{\sqrt{a}+a+1+a\sqrt{a}+a-a-a\sqrt{a}-\sqrt{a}}{a+\sqrt{a}+1}\)
\(=\dfrac{a+1}{a+\sqrt{a}+1}\)
(^~^)
\(\dfrac{a+1}{a+\sqrt{a}+1}=\dfrac{\sqrt{6}}{1+\sqrt{6}}\)
<=> \(a+\sqrt{6}a+1+\sqrt{6}=\sqrt{6}a+\sqrt{6a}+\sqrt{6}\)
<=> \(a-\sqrt{6a}+1=0\left(1\right)\)
<=> \(\left(\sqrt{a}-\sqrt{2+\sqrt{3}}\right)\left(\sqrt{a}-\sqrt{2-\sqrt{3}}\right)=0\)
<=> \(\left[{}\begin{matrix}a=2+\sqrt{3}\\a=2-\sqrt{3}\end{matrix}\right.\)
a)\(x^3+ax+bx+6⋮\left(x-1\right)\)
b)\(x^4+ax^3+bx^2+5x+1⋮\left(x+1\right)^2\)
c)\(^{x^4+3x^3+ax^2+bx+5⋮\left(x-2\right)^2}\)
d)\(x^4+10x^3+ax^2+bx+7⋮\left(x+2\right)^2\)
e)\(x^4+ax^3+5x^2+bx+1⋮x-1\)
Cho a+b+c=0.tính\(\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
Gợi ý thôi ạ ^^! (ai rảnh thì giải luôn cũng được ^^!)
Giải pt: \(3\sqrt[3]{2x+1}+\sqrt{1-x}-4=0\)
Thank you!!!
đkxđ: \(x\le1\)
đặt \(y=\sqrt[3]{2x+1}\)
\(\Rightarrow x=\dfrac{y^3+1}{2}\)
PT\(\Leftrightarrow3y+\sqrt{1-\dfrac{y^3-1}{2}}=4\)
\(\Leftrightarrow4-3y=\sqrt{\dfrac{3-y^3}{2}}đkps:y\le\dfrac{4}{3}\)
\(\Leftrightarrow16-24y+9y^2=\dfrac{3-y^3}{2}\)
\(\Leftrightarrow y^3+18y^2-48y+29=0\)
ta thấy tổng cá hệ số bằng 0 nên pt có 1 no =1
\(\Leftrightarrow\left(y-1\right)\left(y^2+19y-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\left(n\right)\\y=\dfrac{-19+3\sqrt{53}}{2}\left(l\right)\\y=\dfrac{-19-3\sqrt{53}}{2}\left(l\right)\end{matrix}\right.\)
tại y=1 thì x=0 (t/m)
vậy pt có 1 no x=0
\(27\left(2x+1\right)=\left(4-\sqrt{1-x}\right)\left(16+4\sqrt{1-x}+1-x\right)\)
nha6n vao2
\(54x+27=-\sqrt{1-x}\left(17-x\right)-4+4x\)
\(\left(50x+31\right)^2=\left(1-x\right)\left(17-x\right)^2\)
khai triển , giải pt bậc 3, loại nghiệm , kl
Dk: X<=1
Help me,please
1)\(\sqrt{x-2+2\sqrt{x-3}}\)=\(\sqrt{(x-3)+2\sqrt{x-3}+1}=\sqrt{(\sqrt{x-3}+1)^2}=\sqrt{x-3}+1 \)
2)\(\sqrt{x-1-2\sqrt{x-2}}=\sqrt{x-2-2\sqrt{x-2}+1}=\sqrt{(\sqrt{x-2}-1)^2}=\sqrt{x-2}-1\)
Help me, please😣
1. ĐK: x>3\(\sqrt{x-2+2\sqrt{x-3}}=\sqrt{\left(x-3\right)+2\sqrt{x-3}+1}=\sqrt{\left(\sqrt{x-3}+1\right)^2}=\sqrt{x-3}+1\\ \)2( Tương tự) Bớt 1 thêm 1
c) \(\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\left(6-2\sqrt{5}\right)\)
d) \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
c) \(\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\left(6-2\sqrt{5}\right)\)
= \(\left(\sqrt{5}+1\right)\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)\left(5-2\sqrt{5}+1\right)\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{6+2\sqrt{5}}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{5+2\sqrt{5}+1}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{\left(\sqrt{5}+1\right)^2}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)
= \(4.4=16\)
d) \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{1+2+5+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{2}+\sqrt{1}+\sqrt{5}\right)^2}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}\)
= \(1\)
a) \(\sqrt{\dfrac{2-\sqrt{3}}{2}}+\dfrac{1-\sqrt{3}}{2}\)
b) \(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{15}}+2\sqrt{5}-\sqrt{3}\)
a) \(\sqrt{\dfrac{2-\sqrt{3}}{2}}+\dfrac{1-\sqrt{3}}{2}\)
= \(\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)
= \(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}+\dfrac{1-\sqrt{3}}{2}\)
= \(\dfrac{\sqrt{3}-1+1-\sqrt{3}}{2}\)
= 0
b) \(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{15}}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18+20+3+2\sqrt{54}-2\sqrt{360}-2\sqrt{60}}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{\left(\sqrt{18}-\sqrt{20}+\sqrt{3}\right)^2}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18}-2\sqrt{5}+\sqrt{3}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18}\)
1.Cho đa giác đều 91 đỉnh. Mỗi đỉnh của đa giác được tô bởi một trong hai màu xanh hoặc đỏ. Chứng minh luôn tìm được 3 trong 91 đỉnh của đa giác thỏa mãn: 3 đỉnh cùng màu và là 3 đỉnh của một tam giác cân có ít nhất một góc nhỏ hơn\(60^0\)
CÓ PHẢI BẠN ĐANG LUYỆN ĐỀ THI TỈNH KO VẬY DẠNG TOÁN NÀY THƯỜNG XUẤT HIỆN TRONG CÂU CUỐI CỦA CÁC ĐỀ THI TỈNH . MK THẤY KHÓ TOÀN BỎ QUA
HELP ME!
a) \(\sqrt{12}-4\sqrt{75}-3\sqrt{27}+5\sqrt{48}\)
= \(2\sqrt{3}-20\sqrt{3}-9\sqrt{3}+20\sqrt{3}\)
= \(2\sqrt{3}-9\sqrt{3}\)
= \(-7\sqrt{3}\)
b)\(\sqrt{\left(1-2\sqrt{7}\right)^2}+\sqrt{8+2\sqrt{7}}\)
= \(\sqrt{\left(1-2\sqrt{7}\right)^2}+\sqrt{1+2\sqrt{7}+7}\)
= \(2\sqrt{7}-1+\sqrt{\left(1+\sqrt{7}\right)^2}\)
= \(2\sqrt{7}-1+1+\sqrt{7}\)
= \(3\sqrt{7}\)
c) \(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\)
= \(\dfrac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\dfrac{1-\sqrt{3}}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\)
= \(\dfrac{1+\sqrt{3}-1+\sqrt{3}}{1-3}\)
= \(\dfrac{2\sqrt{3}}{-2}=-\sqrt{3}\)