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Bài 4:
Ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
1) \(x-2y=3\Rightarrow\hept{\begin{cases}x=3+2y\\y=\frac{x-3}{2}\end{cases}}\)
\(\Rightarrow A=2x\left(x+2y-3\right)-y\left(6x-3y-10\right)+x-7+\left(x-3y\right)^2\)
\(=2x^2+4xy-6x-6xy+3y^2+10y+x-7+x^2-6xy+9y^2\)
\(=3x^2+12y^2-8xy-5x+10y-7\)
\(=3.\left(3+2y\right)^2+12y^2-8\left(3+2y\right).y-5\left(3+2y\right)+10y-7\)
\(=3\left(9+12y+4y^2\right)+12y^2-8\left(3y+2y^2\right)-15-10y+10y-7\)
\(=27+36y+12y^2+12y^2-24y-16y^2-15-10y+10y-7\)
\(=8y^2+12y+5\)
\(M=\left(x^2-2x+1\right)\left(1+2x\right)-\left(x^2+2x+1\right)\left(1-3x\right)-\left(3-6x\right)\left(x^2+3x+2\right)\)
\(=x^2+2x^3-2x-4x^2+1+2x-x^2+3x^8-2x+6x^2-1+3x-3x^2-9x-6+6x^8\)\(+18x^2+12x=11x^3+17x^2+4x-6\)
2) \(P=\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1=8.\left(\dfrac{1}{2}\right)^3+1=8.\dfrac{1}{8}+1=2\)
\(Q=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3=1^3+27.\left(\dfrac{1}{3}\right)^3=1+27.\dfrac{1}{27}=2\)
3) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow-24x^2+2x+2+24x^2-64x+10=-50\)
\(\Leftrightarrow-62x=-62\Leftrightarrow x=1\)
1)
a) \(\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(=6x^2+7x-5+12x^2+5x-2\)
\(=18x^2+12x-7\)
Vì \(\left|x\right|=2\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
+ Nếu: \(x=2\Rightarrow Bt=18.2^2+12.2-7=89\)
+ Nếu: \(x=-2\Rightarrow Bt=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)
b) Ta có: Tại x=-1/5 , y=-5 thì
\(Bt=25.\left(-\frac{1}{5}\right)^2-2.\left(-\frac{1}{5}\right).\left(-5\right)+\frac{1}{5}.\left(-5\right)^2\)
\(=1-2+5=4\)
Bài 2:
a: \(\Leftrightarrow4x^2-14x+10x-35-\left(4x+3\right)^2=16\)
\(\Leftrightarrow4x^2-4x-35-16x^2-24x-9-16=0\)
\(\Leftrightarrow-12x^2-28x-60=0\)
\(\Leftrightarrow3x^2+7x+15=0\)
\(\text{Δ}=7^2-4\cdot3\cdot15=-131< 0\)
Do đó: Phương trình vô nghiệm
b: Ta có: \(\left(8x^2+3\right)\left(8x^2-3\right)-\left(8x^2-1\right)^2=22\)
\(\Leftrightarrow64x^4-9-64x^4+16x^2-1=22\)
\(\Leftrightarrow16x^2=32\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: Ta có: \(49x^2+14x+1=0\)
=>\(\left(7x+1\right)^2=0\)
hay x=-1/7
1)
Ta có : \(x-3y=5\Rightarrow x=3y+5\)
Thay vào biểu thức A ta được :
\(A=\left(3y+5\right)\left(3y+5-9y+1\right)+3y\left(3y+5+3y-1\right)-2\)
\(=\left(3y+5\right)\left(-6y+6\right)+3y\left(6y+4\right)-2\)
\(=3y\left(-6y+6\right)+5\left(-6y+6\right)+18y^2+12y-2\)
\(=-18y^2+18y-30y+30+18y^2+12y-2\)
\(=30-2=28\)
Vậy : \(A=28\) khi \(x-3y=5\)
1
\(A=x\left(x-9y+1\right)+3y\left(x+3y-1\right)-2\)
\(A=x^2-9xy+x+3xy+9y^2-3y-2\)
\(A=x^2-6xy+9y^2+x-3y-2\)
\(A=\left(x-3y\right)^2+\left(x-2y\right)-2\)
\(A=25-5-2=18\)
bạn kia lm sai r thì phải.nếu đúng thì cho sorry