\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

b. Ta có ∠GHE=360°-(90°+70°+60°)=140°

mà ∠GHE+x=180°⇒x=180°-140°=40°

c. Ta có 2x=360°-(65°+95°)=200°⇒x=200°:2=100°

d. Ta có ∠LKJ=180°-120°=60°

⇒x=360°-(95°+120°+60°)=85°

a.

\(O=AC\cap BD\Rightarrow O\in BD\in\left(SBD\right)\) \(\Rightarrow SO\in\left(SBD\right)\)

\(\left\{{}\begin{matrix}SO\perp\left(ABCD\right)\Rightarrow SO\perp AC\\AC\perp BD\left(gt\right)\end{matrix}\right.\) \(\Rightarrow AC\perp\left(SBD\right)\Rightarrow AC\perp SD\)

b.

O là trung điểm AC, H là trung điểm AB \(\Rightarrow\) OH là đường trung bình tam giác ABC

\(\Rightarrow OH||BC\Rightarrow OH\perp AB\Rightarrow OH\perp CD\) (1)

Mà \(SO\perp\left(ABCD\right)\Rightarrow SO\perp CD\) (2)

(1);(2) \(\Rightarrow CD\perp\left(SHO\right)\)

c.

Theo cmt trên \(OH||BC\Rightarrow OH||AD\)

\(\Rightarrow\widehat{\left(OH;SD\right)}=\widehat{\left(AD;SD\right)}=\widehat{SDA}\)

\(AC=2a\sqrt{2}\Rightarrow OA=a\sqrt{2}\Rightarrow SA=SB=SC=SD=\sqrt{SO^2+OA^2}=a\sqrt{3}\)

Áp dụng định lý hàm cosin trong tam giác SAD:

\(cos\widehat{SDA}=\dfrac{SD^2+AD^2-SA^2}{2SD.AD}=\dfrac{\sqrt{3}}{3}\)

\(\Rightarrow\widehat{SDA}=...\)

d.

Gọi E là trung điểm SB \(\Rightarrow HE\) là đường trung bình tam giác SAB 

\(\Rightarrow HE||SA\Rightarrow\widehat{\left(HK;SA\right)}=\widehat{\left(HK;HE\right)}=\widehat{KHE}\)

\(SK=\dfrac{1}{3}SC=\dfrac{a\sqrt{3}}{3}\) ; \(SE=\dfrac{1}{2}SB=\dfrac{a\sqrt{3}}{2}\) ; \(EH=\dfrac{1}{2}SA=\dfrac{a\sqrt{3}}{2}\)

\(cos\widehat{BSC}=\dfrac{SB^2+SC^2-BC^2}{2SB.SC}=\dfrac{1}{3}\)

\(\Rightarrow EK=\sqrt{SE^2+SK^2-2SE.SK.cos\widehat{BSC}}=\dfrac{a\sqrt{3}}{2}\)

Từ K kẻ KF song song SO \(\Rightarrow KF\perp\left(ABCD\right)\Rightarrow KF\perp HF\)

\(\dfrac{KF}{SO}=\dfrac{CK}{CS}=\dfrac{2}{3}\Rightarrow KF=\dfrac{2a}{3}\)

\(\dfrac{OF}{OC}=\dfrac{SK}{SC}=\dfrac{1}{3}\Rightarrow OF=\dfrac{1}{3}OC\Rightarrow AF=\dfrac{4}{3}OC=\dfrac{2}{3}AC=\dfrac{4a\sqrt{2}}{3}\)

\(\Rightarrow HF=\sqrt{AH^2+AF^2-2AH.AF.cos45^0}=\dfrac{a\sqrt{17}}{3}\)

\(\Rightarrow HK=\sqrt{HF^2+KF^2}=\dfrac{a\sqrt{21}}{3}\)

\(\Rightarrow cos\widehat{KHE}=\dfrac{HK^2+EH^2-EK^2}{2HK.EH}=\dfrac{\sqrt{7}}{3}\)

a: \(x^3y+x-y-1\)

\(=\left(x^3y-y\right)+\left(x-1\right)\)

\(=y\left(x^3-1\right)+\left(x-1\right)\)

\(=y\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2y+xy+y+1\right)\)

b: \(x^2\left(x-2\right)+4\left(2-x\right)\)

\(=x^2\left(x-2\right)-4\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2-4\right)\)

\(=\left(x-2\right)\cdot\left(x-2\right)\left(x+2\right)=\left(x+2\right)\cdot\left(x-2\right)^2\)

c: \(x^3-x^2-20x\)

\(=x\cdot x^2-x\cdot x-x\cdot20\)

\(=x\left(x^2-x-20\right)\)

\(=x\left(x^2-5x+4x-20\right)\)

\(=x\left[x\left(x-5\right)+4\left(x-5\right)\right]\)

\(=x\left(x-5\right)\left(x+4\right)\)

d: \(\left(x^2+1\right)^2-\left(x+1\right)^2\)

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

\(=\left(x^2+x+2\right)\left(x^2-x\right)\)

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

ảnh kia nhiều người lắm like thế :)

thì ai cũng chịu mà

:))))