= 0

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc đề của bạn dễ hiểu hơn nhé.

a: =(căn a-3)^2-b^2

=(căn a-3-b)(căn a-3+b)

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

c: \(x-7\sqrt{x}+12=x-3\sqrt{x}-4\sqrt{x}+12=\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)\)

d: x*căn x-64

=(căn x)^3-4^3

=(căn x-4)(x+4căn x+16)

\(a-6\sqrt{a}+9-b^2\\ =\left(\sqrt{a}+3\right)^2-b^2\\ =\left(\sqrt{a}+3-b\right)\left(\sqrt{a}+3+b\right)\)

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

\(x-7\sqrt{x}+12\\ =x-4\sqrt{x}-3\sqrt{x}+12\\ =\sqrt{x}\left(\sqrt{x}-4\right)-3\left(\sqrt{x}-4\right)\\ =\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\)

\(x\sqrt{x}+64\\ =\sqrt{x^3}+4^3\\ =\left(\sqrt{x}\right)^3+4^3\\ =\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)\)

e) \(\left(\dfrac{-1}{3}\right)\cdot\dfrac{5}{7}=\dfrac{\left(-1\right)\cdot5}{3\cdot7}=\dfrac{-5}{21}\)

f) \(\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}\cdot\dfrac{4}{3}=\dfrac{8}{21}\)

\(e.\left(\dfrac{-1}{3}\right).\dfrac{5}{7}=\dfrac{\left(-1\right).5}{3.7}=\dfrac{-5}{21}\\ f.\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}.\dfrac{4}{3}=\dfrac{2.4}{7.3}=\dfrac{8}{21}\)

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

\(\Leftrightarrow-x^2-x^2+3x-2x-3x-8x=-6+6\)

\(\Leftrightarrow-2x^2-10x=0\)

\(\Leftrightarrow-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;5\right\}\)

có, đúng rồi

em cảm ơn ạ

Nhân chia trước,cộng trừ sau nhé!:D

\(a,=\left(2\sqrt{6}-4\sqrt{3}\right)\sqrt{6}+12\sqrt{2}=12-12\sqrt{2}+12\sqrt{2}=12\\ b,=\dfrac{6\left(3-\sqrt{3}\right)}{6}+\sqrt{3}=3-\sqrt{3}+\sqrt{3}=3\)

a.

\(2x-x^2+7=-\left(x^2-2x+1\right)+8=-\left(x-1\right)^2+8\le8\)

\(\Rightarrow2+\sqrt{2x-x^2+7}\le2+\sqrt{8}=2+2\sqrt{2}\)

\(\Rightarrow\dfrac{3}{2+\sqrt{2x-x^2+7}}\ge\dfrac{3}{2+2\sqrt{2}}=\dfrac{3\sqrt{2}-3}{2}\)

\(A_{min}=\dfrac{3\sqrt{2}-3}{2}\) khi \(x=1\)

b. ĐKXĐ: \(x\le1\)

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

\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}\right)+\dfrac{3}{2}\)

\(B=-\left(\sqrt{1-x}-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}\le\dfrac{3}{2}\)

\(B_{max}=\dfrac{3}{2}\) khi\(x=\dfrac{1}{2}\)