c,C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\left(x\ge1\right)\)

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

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

TH1: \(\sqrt{x-1}< 1\) hay \(1\le x< 2\)

Từ (1)=>C= \(\sqrt{x-1}+1+1-\sqrt{x-1}\)=2

TH2: \(\sqrt{x-1}\ge1\) hay \(x\ge2\)

Từ (1) =>C=\(\sqrt{x-1}+1+\sqrt{x-1}-1\)=\(2\sqrt{x-1}\)

d, D=\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{13+30\sqrt{2}+\sqrt{8+2\sqrt{8}+1}}=\sqrt{13+30\sqrt{2}+\sqrt{\left(\sqrt{8}+1\right)^2}}\)

=\(\sqrt{13+30\sqrt{2}+\sqrt{8}+1}=\sqrt{14+30\sqrt{2}+2\sqrt{2}}\)

=\(\sqrt{14+32\sqrt{2}}\)

a)\(\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)

b)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)

\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)

\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)

\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)

\(B=3-\sqrt{x}-\sqrt{x}+3-6\)

\(B=-2\sqrt{x}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3}{\sqrt{x}-6}\)

Sửa đề nha: \(\sqrt{x^3-1}\) thành \(\sqrt{x^3}-1\)

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

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

\(B=\frac{\left(x+\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)

b/ Để B= 3\(\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=16\)

thi vào 10 xong rồi vẫn chịu khó giải bài lớp 9 hở

\(A=\frac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x+16}{\sqrt{x}+3}\)

+ \(A=\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}\) \(=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{25}{\sqrt{x}+3}}-6=10-6=4\)

Dấu "=" \(\Leftrightarrow\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}+3=5\Leftrightarrow x=4\)

Vậy \(A=\frac{x+16}{\sqrt{x}+3}\)

Min A = 4 \(\Leftrightarrow x=4\)

cho hỏi là mẫu biểu thức A là\(\sqrt{x}-3\) hay\(\sqrt{x-3}\)

là\(\sqrt{x}-3\)mình ghi nhầm

\(P=\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{2}{\sqrt{x}-1}\)

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

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

Do \(x+\sqrt{x}+1=x+\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow P=\frac{2}{x+\sqrt{x}+1}>0\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaaI2a % GaeyOeI0IaaGOmaiaadIhacqGHsisldaGcaaqaaiaaiMdacqGHsisl % caaI2aGaamiEaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaabe % aakmaabmaabaGaamiEaiabgYda8iaaiodaaiaawIcacaGLPaaaaeaa % cqGH9aqpcaaI2aGaeyOeI0IaaGOmaiaadIhacqGHsisldaGcaaqaam % aabmaabaGaaG4maiabgkHiTiaadIhaaiaawIcacaGLPaaadaahaaWc % beqaaiaaikdaaaaabeaaaOqaaiabg2da9iaaiAdacqGHsislcaaIYa % GaamiEaiabgkHiTmaaemaabaGaaG4maiabgkHiTiaadIhaaiaawEa7 % caGLiWoaaeaacqGH9aqpcaaI2aGaeyOeI0IaaGOmaiaadIhacqGHRa % WkcaaIZaGaeyOeI0IaamiEaaqaaiabg2da9iaaiMdacqGHsislcaaI % ZaGaamiEaaqaamaalaaabaGaaG4maiabgkHiTmaakaaabaGaamiEaa % WcbeaaaOqaaiaadIhacqGHsislcaaI5aaaamaabmaabaGaamiEaiab % gwMiZkaaicdacaGGSaGaamiEaiabgcMi5kaaiMdaaiaawIcacaGLPa % aaaeaacqGH9aqpdaWcaaqaaiabgkHiTmaabmaabaWaaOaaaeaacaWG % 4baaleqaaOGaeyOeI0IaaG4maaGaayjkaiaawMcaaaqaamaabmaaba % WaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaG4maaGaayjkaiaawMca % amaabmaabaWaaOaaaeaacaWG4baaleqaaOGaey4kaSIaaG4maaGaay % jkaiaawMcaaaaaaeaacqGH9aqpdaWcaaqaaiabgkHiTiaaigdaaeaa % daGcaaqaaiaadIhaaSqabaGccqGHRaWkcaaIZaaaaaqaamaalaaaba % GaamiEaiabgkHiTiaaiwdadaGcaaqaaiaadIhaaSqabaGccqGHRaWk % caaI2aaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaG4maaaada % qadaqaaiaadIhacqGHLjYScaaIWaGaaiilaiaadIhacqGHGjsUcaaI % 5aaacaGLOaGaayzkaaaabaGaeyypa0ZaaSaaaeaacaWG4bGaeyOeI0 % IaaGOmamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaiodadaGcaaqa % aiaadIhaaSqabaGccqGHRaWkcaaI2aaabaWaaOaaaeaacaWG4baale % qaaOGaeyOeI0IaaG4maaaaaeaacqGH9aqpdaWcaaqaamaakaaabaGa % amiEaaWcbeaakmaabmaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0 % IaaGOmaaGaayjkaiaawMcaaiabgkHiTiaaiodadaqadaqaamaakaaa % baGaamiEaaWcbeaakiabgkHiTiaaikdaaiaawIcacaGLPaaaaeaada % GcaaqaaiaadIhaaSqabaGccqGHsislcaaIZaaaaaqaaiabg2da9maa % laaabaWaaeWaaeaadaGcaaqaaiaadIhaaSqabaGccqGHsislcaaIYa % aacaGLOaGaayzkaaWaaeWaaeaadaGcaaqaaiaadIhaaSqabaGccqGH % sislcaaIZaaacaGLOaGaayzkaaaabaWaaOaaaeaacaWG4baaleqaaO % GaeyOeI0IaaG4maaaaaeaacqGH9aqpdaGcaaqaaiaadIhaaSqabaGc % cqGHsislcaaIYaaaaaa!C78C! \begin{array}{l} 6 - 2x - \sqrt {9 - 6x + {x^2}} \left( {x < 3} \right)\\ = 6 - 2x - \sqrt {{{\left( {3 - x} \right)}^2}} \\ = 6 - 2x - \left| {3 - x} \right|\\ = 6 - 2x + 3 - x\\ = 9 - 3x\\ \dfrac{{3 - \sqrt x }}{{x - 9}}\left( {x \ge 0,x \ne 9} \right)\\ = \dfrac{{ - \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{ - 1}}{{\sqrt x + 3}}\\ \dfrac{{x - 5\sqrt x + 6}}{{\sqrt x - 3}}\left( {x \ge 0,x \ne 9} \right)\\ = \dfrac{{x - 2\sqrt x - 3\sqrt x + 6}}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - 3\left( {\sqrt x - 2} \right)}}{{\sqrt x - 3}}\\ = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\ = \sqrt x - 2 \end{array}\)

\(6-2x-\sqrt{9-6x+x^2}\)

= \(6-2x-\sqrt{\left(3-x\right)^2}\)

= \(\left\{{}\begin{matrix}6-2x-3+x\\6-2x+3-x\end{matrix}\right.\)

= \(\left\{{}\begin{matrix}3-x\\9-3x\end{matrix}\right.\)

\(\frac{3-\sqrt{x}}{x-9}\)

=\(\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(x-3\right)}\)

= \(\frac{-1}{\sqrt{x}+3}\)