Kết quả tìm kiếm

Rút gọn: $a)\sqrt{6+2\sqrt{5}}-\sqrt{21-4\sqrt{5}} \\=\sqrt{5+2\sqrt{5}+1}-\sqrt{20-4\sqrt{5}+1} \\=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(2\sqrt{5}-1)^2} \\=\sqrt{5}+1-2\sqrt{5}+1 \\=2-\sqrt{5} \\b)2\sqrt{8-2\sqrt{7}}+\sqrt{64+6\sqrt{7}} \\=2\sqrt{7-2\sqrt{7}+1}+\sqrt{63+6\sqrt{7}+1}...

EXERCISE I: Put the verbs in brackets in correct tenses. 1/ If I (have) ____had____ a typewriter, I could type it myself . HAD 2/ If I had known that you were in hospital , I (visit) _____would have visited___ you. 3/ You could make much progress if you (attend) ____attended____ class...

$2)\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}} \\=\sqrt{9-6\sqrt 6+6}+\sqrt{24-12\sqrt 6+9} \\=\sqrt{(3-\sqrt 6)^2}+\sqrt{(2\sqrt 6-3)^2} \\=3-\sqrt 6+2\sqrt 6-3 \\=\sqrt 6$

a) ĐK: $x\geq 1;x\neq 3$ $P=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}=\dfrac{(x-3)(\sqrt{x-1}+\sqrt{2})}{x-1-2}=\sqrt{x-1}+\sqrt{2}$ $b) \sqrt{x-1}\geq 0\Rightarrow \sqrt{x-1}+\sqrt 2\geq \sqrt 2$ hay $P\geq \sqrt 2$ Dấu '=' xảy ra khi $x=1$ Vậy...

1) a) ĐK: $x>0;x\neq 1$ $a)M=(\dfrac{\sqrt a}2-\dfrac1{2\sqrt a})(\dfrac{a-\sqrt a}{\sqrt a+1}-\dfrac{a+\sqrt a}{\sqrt a-1}) \\=\dfrac{(\sqrt a+1)(\sqrt a-1)}{2\sqrt a}.\dfrac{\sqrt a[(\sqrt a-1)^2-(\sqrt a+1)^2]}{(\sqrt a+1)(\sqrt a-1)} \\=\dfrac{a-2\sqrt a+1-a-2\sqrt a-1}{2} \\=-2\sqrt{a}$...

a) $BD$ là đường phân giác $\Rightarrow \dfrac{AD}{CD}=\dfrac{AB}{BC}=\dfrac 45$ $\Rightarrow \dfrac{AD}{6}=\dfrac 4{9}\Rightarrow AD=\dfrac 83(cm)\Rightarrow CD=\dfrac{10}3cm$ $CE$ là đường phân giác $\Rightarrow \dfrac{AE}{EB}=\dfrac{AC}{BC}=\dfrac 35$ $\Rightarrow \dfrac{AE}{8}=\dfrac...

1) $0<A\leq 5\Rightarrow A\in \left \{ 1;2;3;4;5 \right \}$ 2) a) Ta có: VT $=\sqrt{x^2-6x+9}+\sqrt{x^2+8x+16}=\sqrt{(x-3)^2}+\sqrt{(x+4)^2}=|x-3|+|x+4|=|3-x|+|x+4|\geq |3-x+x+4|=7$ Dấu '=' xảy ra khi $(3-x)(x+4)\geq 0\Leftrightarrow -4\leq x\leq 3$ $\Rightarrow \cdots$

$a)4x^2y^2 - (x^ 2+ y^2 - 4z^2)^2=(x+y-2z)(x+y+2z)(x-y+2z)(y-x+2z) \\b)a^3 - 9(3c^3 - 2abc) + 8b^3=(a+2b-3c)(a^2+4b^2+9c^2-2ab+6bc+3ac) \\c)bc(b + c) - ca(c + a) + ab(b - a)=(b-a)(a+c)(b+c) \\d)a(ab - b^2) + b(bc - c^2) + c(ac - a^2)=(a-b)(b-c)(a-c) \\e)xy(x^2 - y^2) + yz(x^2 - y^2) + zx(z^2 -...

$a)\dfrac{x+1}{2x+2}-\dfrac{x-1}{2x+2}+\dfrac{2x}{(1-x)(x-1)}=0 \ \ \ (x\neq \pm 1) \\\Leftrightarrow \dfrac{1}{x+1}-\dfrac{2x}{(x-1)^2}=0 \\\Leftrightarrow (x-1)^2-2x(x+1)=0 \\\Leftrightarrow x^2+4x-1=0 \\\Leftrightarrow (x+2)^2=5$ \\\Leftrightarrow |x+2|=\sqrt 5 \\\Leftrightarrow x+2=\pm...

$1)x^6-y^6=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2) \\2)x^3 \color{red}{+} 9x^2 + 11x - 21=(x-1)(x+3)(x+7) \\3)y(x-2z)^2 + 8xyz + x(y-2z)^2 - 2z(x+y)^2= (x + y)(x - 2z)(y - 2z) \\4)x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2 + 2xyz=(x+y) (y+z)(z+x) \\5)( x^2 + y^2 )^3 + (z^2 - x^2 )^3 - ( y^2 + z^2...

2) $OB^2=AB^2-OA^2=16\Rightarrow OB=4(cm)$ $AB^2=OB.BD\Rightarrow BD=\dfrac{AB^2}{OB}=13(cm)$ $\Rightarrow AD=BD^2-AB^3=117\Rightarrow AD=3\sqrt{13}(cm)$ $OA.OC=OD^2\Rightarrow OC=\dfrac{OD^2}{OA}=13,5(cm)\Rightarrow AC=19,5cm$ $\Rightarrow CD^2=AC^2-AD^2=263,25(cm)\Rightarrow...

chắc nhầm chỗ đấy nhưng bình phương lên vẫn vậy ĐKXĐ thiếu $x^2-x-1\leq 0$