d)x=1/7=0.142857143

  x=1

x= \(\frac{1}{7}\)= 0.142857143

x = 1

a b x x Z G H A

Kéo dài AZ sao cho AZ cắt đường thẳng b tại G.

Ta có : a//b => ^aAZ=^ZGH ( so le trong)

=> x=^ZGH

Xét tam giác ZGH : ^GZH=90^o

=> ^ZGH + ^ZHG=90^o ( đ/lý tổng 3 góc tam giác trong 1 tam giác vuông )

Có ^ZGH=^ZHG ( = x)

=> ^ZGH=^ZHG =45^o

=> x=45^o

\(\left(2x-1\right)\left(3x-2\right)-\left(6x-1\right)\left(x-3\right)=17\)

\(\Rightarrow\left(6x^2-4x-3x+2\right)-\left(6x^2-18x-x+3\right)=17\)

\(\Rightarrow6x^2-4x-3x+2-6x^2+18x+x-3=17\)

\(\Rightarrow\left(6x^2-6x^2\right)+\left(18x-4x-3x+x\right)-\left(3+2\right)=17\)

\(\Rightarrow12x-1=17\)

\(\Rightarrow12x=18\)

\(\Rightarrow x=\frac{3}{2}\)

6x² - 4x - 3x + 2 - 6x² - 18x - x + 3 = 17 

=>>> -27x + 5 = 17 =>>> -27x = 12 =>>> x = -4/7

Vậy: x = -4/7

giúp mk với

Ta có : \(A=6x^3-x\left(x+2\right)+4\left(x+3\right)=6x^3-x^2-2x+4x+12\)

\(=6x^3-x^2+2x+12\)

\(B=-x\left(x+1\right)-\left(4-3x\right)+x^2\left(x-2\right)=-x^2-x-4+3x+x^3-2x^2\)

\(=-3x^2+2x-4+x^3\)

a, \(A+B=6x^3-x^2+2x+12-3x^2+2x-4+x^3\)

\(=7x^3-4x^2+4x+8\)

b, \(B-A=-3x^2+2x-4+x^3-6x^3+x^2-2x-12\)

\(=-2x^2-16-5x^3\)

\(\left(x+1\right)^2=x^2+2\cdot x\cdot1+1^2=x^2+2x+1=VP\left(đpcm\right)\)

\(P\left(x\right)=x^2+2x+4\)

\(\Delta=b^2-4ac=2^2-4\cdot1\cdot4=4-16=-12\)

\(\Delta< 0\)=> Đa thức vô nghiệm ( đpcm ) 

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

=>  \(x^2+2x+1=x^2+2x+1\left(\text{đ}pcm\right)\)

Ta có : \(P\left(x\right)=x^2+2x+4=0\)

\(\hept{\begin{cases}x^2\ge0\\2x\ge0\\4>0\end{cases}\Rightarrow vonghiem}\)

\(A=\frac{9}{10}\times\frac{10}{11}\times\frac{11}{12}\times...\times\frac{99}{100}\)

\(A=\frac{9\times10\times11\times...99}{10\times11\times12\times...\times100}\)

\(\Rightarrow A=\frac{9}{100}\)

A B x C y 95

Ta kéo Ax cắt CB tại H ( mình quyên đặt ...hihi) 

Vì Ax // Cy mà \(\widehat{C}\)= 95⁰ \(\Rightarrow\)\(\widehat{AHB}\) = 95⁰, \(\widehat{B}\)= 30⁰  \(\Rightarrow\)\(\widehat{HAB}\) = 180⁰ - 95⁰ - 30⁰ = 55

\(\Rightarrow\)\(\widehat{A}\)= 180⁰ - 55⁰( kề bù)  =  125⁰  

Vậy: \(\widehat{A}\)= 125⁰

a) M = x² + 4y² + 4xy = x² + 2xy + 2xy + 4y² = x(x + 2y) + 2y(x + 2y) = (x + 2y)² \(\ge\)0 \(\forall\)x;y

b) x = 4y, ta có: M = 9 

<=> (4y + 2y)² = 9

<=> 36y² = 9

<=> y² = 1/4

<=> \(\orbr{\begin{cases}y=\frac{1}{2}\\y=-\frac{1}{2}\end{cases}}\)

Với y = 1/2 => x = 4.1/2 = 2

y = -1/2 => x = 4. (-1/2) = -2

\(M=x^2+4y^2+4xy\)

\(\Rightarrow M=x^2+\left(2y\right)^2+2xy+2xy\)

\(\Rightarrow M=x^2+\left(2y\right)^2+\left(2xy\right)^2\ge0\forall x;y\)

\(\Rightarrow M\ge0\forall x;y\)