Câu 4: \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\\ \)

Đặt y=x^2-5x được:\(\left(y-6\right)\left(y+6\right)=y^2-36\ge-36\)

Vậy GTNN của a là -36 tại x=0 hoặc x=5

\(x+y=2\Rightarrow x=2-y\\ 3x^2+y^2=3\left(2-y\right)^2+y^2=12-12y+3y^2+y^2=\left(2y-3\right)^2+3\ge3\)

Vậy GTNN của A là 3 tại y=3/2;x=1/2

\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.\left(\dfrac{c}{z}+\dfrac{b}{y}+\dfrac{a}{x}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.0=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)

10)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}=0\Rightarrow xy+yz+xz=0\Leftrightarrow\left(xy\right)^3+\left(yz\right)^3+\left(xz\right)^3=3xy.yz.xz\Leftrightarrow\left(xy\right)^3+\left(yz\right)^3+\left(xz\right)^3=3x^2y^2z^2\\ \dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=\dfrac{\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3}{x^2y^2z^2}=\dfrac{3x^2y^2z^2}{x^2y^2z^2}=3\)

5)\(AB+AC=22\Leftrightarrow AB+1.75AB=22\Leftrightarrow AB=8;AC=14\)

\(S_{ABC}=\dfrac{1}{2}.AB.AC.\sin A=\dfrac{1}{2}.8.14.\dfrac{1}{2}=28cm^2\)

4)Gọi hai số đó là a và b

\(a+b=170\Rightarrow a=170-b\\ \dfrac{1}{2}a+\dfrac{2}{5}b=21\Leftrightarrow\dfrac{1}{2}\left(170-b\right)+\dfrac{2}{5}b=21\Leftrightarrow-\dfrac{1}{10}b=-64\Leftrightarrow b=640;a=170-640=-470\)

2)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=[x\left(x+3\right)].\left[\left(x+1\right)\left(x+2\right)\right]=\left(x^2+3x\right)\left(x^2+3x+2\right)\)

Đặt y=\(x^2+3x\) ta được:\(y\left(y+2\right)=y^2+2y=\left(y+1\right)^2-1\)

Vậy GTNN của x(x+1)(x+2)(x+3) là -1

1)\(S=a^2=121\Rightarrow a=11\\P=4a=4.11=44\)

2)\(h2=\dfrac{AB.h1}{BC}=\dfrac{6.5}{4}=7,5\)

3)\(M=\dfrac{500^2}{127^2-123^2}=\dfrac{500^2}{\left(127+123\right).\left(127-123\right)}=\dfrac{4.250^2}{4.250}=250\)

4)\(ĐKXĐ:n\ne0\\ \dfrac{1}{2}+\dfrac{1}{n}>\dfrac{1}{4}+\dfrac{2}{5}\Leftrightarrow\dfrac{1}{n}>\dfrac{3}{20}\Leftrightarrow20>3n\)

Vậy n lớn nhất là 6

5)Bạn bấm máy sẽ ra

7)Gọi số tự nhiên cần tìm là a

a chia 4 dư 3=>a+1 chia hết cho 4(1)

a chia 3 dư 2=>a+1 chia hết cho 3(2)

a chia 2 dư 1=>a+1 chia hết cho 2(3)

Từ (1),(2) và (3) suy ra a+1 thuộc BCNN(2,3,4)=12=>a=12-1=11

8)\(5^{x^2-35}=25^x\Leftrightarrow5^{x^2-35}=5^{2x}\Rightarrow x^2-35=2x\Leftrightarrow x^2-2x-35=0\Leftrightarrow\left(x^2-7x\right)+\left(5x-35\right)=0\Leftrightarrow x\left(x-7\right)+5\left(x-7\right)=0\Leftrightarrow\left(x-7\right)\left(x+5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=7\left(nhan\right)\\x=-5\left(loai\right)\end{matrix}\right.\)

Vậy x=7

9)15